Introduction to Inequalities

Inequality tells us about the relative size of values.

Mathematics is not always about "equals", sometimes we only know that something is greater or less than.

Example: Alex and Billy have a race, and Billy wins!

What do we know?

We don't know how fast they ran, but we do know that Billy was faster than Alex:

Billy was faster than Alex

We can write that down like this:

(Where "b" means how fast Billy was, ">" means "greater than", and "a" means how fast Alex was)

We call things like that inequalities (because they are not "equal")

Greater or Less Than

The two most common inequalities are:

They are easy to remember: the "small" end always points to the smaller number, like this:

Greater Than Symbol: BIG > small

Example: Alex plays in the under 15s soccer. How old is Alex?

We don't know exactly how old Alex is, because it doesn't say "equals"

But we do know "less than 15", so we can write:

Age < 15

The small end points to "Age" because the age is smaller than 15.

... Or Equal To!

We can also have inequalities that include "equals", like:

Example: you must be 13 or older to watch a movie.

The "inequality" is between your age and the age of 13 .

Your age must be "greater than or equal to 13", which is written:

Comparing Values

Practice >, < and = with Compare Numbers to 10

Learn more about Inequalities at Less Than or Greater Than

  • Introduction to Functions

1.1 Functions and Function Notation

1.2 domain and range, 1.3 rates of change and behavior of graphs, 1.4 composition of functions, 1.5 transformation of functions, 1.6 absolute value functions, 1.7 inverse functions.

  • Key Equations
  • Key Concepts

Review Exercises

Practice test.

  • Introduction to Linear Functions
  • 2.1 Linear Functions
  • 2.2 Graphs of Linear Functions
  • 2.3 Modeling with Linear Functions
  • 2.4 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 3.1 Complex Numbers
  • 3.2 Quadratic Functions
  • 3.3 Power Functions and Polynomial Functions
  • 3.4 Graphs of Polynomial Functions
  • 3.5 Dividing Polynomials
  • 3.6 Zeros of Polynomial Functions
  • 3.7 Rational Functions
  • 3.8 Inverses and Radical Functions
  • 3.9 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 4.1 Exponential Functions
  • 4.2 Graphs of Exponential Functions
  • 4.3 Logarithmic Functions
  • 4.4 Graphs of Logarithmic Functions
  • 4.5 Logarithmic Properties
  • 4.6 Exponential and Logarithmic Equations
  • 4.7 Exponential and Logarithmic Models
  • 4.8 Fitting Exponential Models to Data
  • Introduction to Trigonometric Functions
  • 5.2 Unit Circle: Sine and Cosine Functions
  • 5.3 The Other Trigonometric Functions
  • 5.4 Right Triangle Trigonometry
  • Introduction to Periodic Functions
  • 6.1 Graphs of the Sine and Cosine Functions
  • 6.2 Graphs of the Other Trigonometric Functions
  • 6.3 Inverse Trigonometric Functions
  • Introduction to Trigonometric Identities and Equations
  • 7.1 Solving Trigonometric Equations with Identities
  • 7.2 Sum and Difference Identities
  • 7.3 Double-Angle, Half-Angle, and Reduction Formulas
  • 7.4 Sum-to-Product and Product-to-Sum Formulas
  • 7.5 Solving Trigonometric Equations
  • 7.6 Modeling with Trigonometric Functions
  • Introduction to Further Applications of Trigonometry
  • 8.1 Non-right Triangles: Law of Sines
  • 8.2 Non-right Triangles: Law of Cosines
  • 8.3 Polar Coordinates
  • 8.4 Polar Coordinates: Graphs
  • 8.5 Polar Form of Complex Numbers
  • 8.6 Parametric Equations
  • 8.7 Parametric Equations: Graphs
  • 8.8 Vectors
  • Introduction to Systems of Equations and Inequalities
  • 9.1 Systems of Linear Equations: Two Variables
  • 9.2 Systems of Linear Equations: Three Variables
  • 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 9.4 Partial Fractions
  • 9.5 Matrices and Matrix Operations
  • 9.6 Solving Systems with Gaussian Elimination
  • 9.7 Solving Systems with Inverses
  • 9.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 10.1 The Ellipse
  • 10.2 The Hyperbola
  • 10.3 The Parabola
  • 10.4 Rotation of Axes
  • 10.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 11.1 Sequences and Their Notations
  • 11.2 Arithmetic Sequences
  • 11.3 Geometric Sequences
  • 11.4 Series and Their Notations
  • 11.5 Counting Principles
  • 11.6 Binomial Theorem
  • 11.7 Probability
  • Introduction to Calculus
  • 12.1 Finding Limits: Numerical and Graphical Approaches
  • 12.2 Finding Limits: Properties of Limits
  • 12.3 Continuity
  • 12.4 Derivatives
  • A | Basic Functions and Identities
  • ⓑ yes. (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)

w = f ( d ) w = f ( d )

g ( 5 ) = 1 g ( 5 ) = 1

m = 8 m = 8

y = f ( x ) = x 3 2 y = f ( x ) = x 3 2

g ( 1 ) = 8 g ( 1 ) = 8

x = 0 x = 0 or x = 2 x = 2

  • ⓐ yes, because each bank account has a single balance at any given time
  • ⓑ no, because several bank account numbers may have the same balance
  • ⓒ no, because the same output may correspond to more than one input.
  • ⓐ Yes, letter grade is a function of percent grade;
  • ⓑ No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.

No, because it does not pass the horizontal line test.

{ − 5 , 0 , 5 , 10 , 15 } { − 5 , 0 , 5 , 10 , 15 }

( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ ) ( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ )

[ − 5 2 , ∞ ) [ − 5 2 , ∞ )

  • ⓐ values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3;
  • ⓑ { x | x ≤ − 2 or − 1 ≤ x < 3 } { x | x ≤ − 2 or − 1 ≤ x < 3 } ;
  • ⓒ ( − ∞ , − 2 ] ∪ [ − 1 , 3 ) ( − ∞ , − 2 ] ∪ [ − 1 , 3 )

domain =[1950,2002] range = [47,000,000,89,000,000]

domain: ( − ∞ , 2 ] ; ( − ∞ , 2 ] ; range: ( − ∞ , 0 ] ( − ∞ , 0 ]

$ 2.84 − $ 2.31 5  years = $ 0.53 5  years = $ 0.106 $ 2.84 − $ 2.31 5  years = $ 0.53 5  years = $ 0.106 per year.

a + 7 a + 7

The local maximum appears to occur at ( − 1 , 28 ) , ( − 1 , 28 ) , and the local minimum occurs at ( 5 , − 80 ) . ( 5 , − 80 ) . The function is increasing on ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) and decreasing on ( − 1 , 5 ) . ( − 1 , 5 ) .

( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2 ( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2

No, the functions are not the same.

A gravitational force is still a force, so a ( G ( r ) ) a ( G ( r ) ) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G ( a ( F ) ) G ( a ( F ) ) does not make sense.

f ( g ( 1 ) ) = f ( 3 ) = 3 f ( g ( 1 ) ) = f ( 3 ) = 3 and g ( f ( 4 ) ) = g ( 1 ) = 3 g ( f ( 4 ) ) = g ( 1 ) = 3

g ( f ( 2 ) ) = g ( 5 ) = 3 g ( f ( 2 ) ) = g ( 5 ) = 3

[ − 4 , 0 ) ∪ ( 0 , ∞ ) [ − 4 , 0 ) ∪ ( 0 , ∞ )

Possible answer:

g ( x ) = 4 + x 2 g ( x ) = 4 + x 2 h ( x ) = 4 3 − x h ( x ) = 4 3 − x f = h ∘ g f = h ∘ g

The graphs of f ( x ) f ( x ) and g ( x ) g ( x ) are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.

g ( x ) = 1 x - 1 + 1 g ( x ) = 1 x - 1 + 1

g ( x ) = − f ( x ) g ( x ) = − f ( x )

h ( x ) = f ( − x ) h ( x ) = f ( − x )

Notice: g ( x ) = f ( − x ) g ( x ) = f ( − x ) looks the same as f ( x ) f ( x ) .

g ( x ) = 3 x - 2 g ( x ) = 3 x - 2

g ( x ) = f ( 1 3 x ) g ( x ) = f ( 1 3 x ) so using the square root function we get g ( x ) = 1 3 x g ( x ) = 1 3 x

| x − 2 | ≤ 3 | x − 2 | ≤ 3

using the variable p p for passing, | p − 80 | ≤ 20 | p − 80 | ≤ 20

f ( x ) = − | x + 2 | + 3 f ( x ) = − | x + 2 | + 3

x = − 1 x = − 1 or x = 2 x = 2

f ( 0 ) = 1 , f ( 0 ) = 1 , so the graph intersects the vertical axis at ( 0 , 1 ) . ( 0 , 1 ) . f ( x ) = 0 f ( x ) = 0 when x = − 5 x = − 5 and x = 1 x = 1 so the graph intersects the horizontal axis at ( − 5 , 0 ) ( − 5 , 0 ) and ( 1 , 0 ) . ( 1 , 0 ) .

- 8 ≤ x ≤ 4 - 8 ≤ x ≤ 4

k ≤ 1 k ≤ 1 or k ≥ 7 ; k ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ 7 , ∞ ) ( − ∞ , 1 ] ∪ [ 7 , ∞ )

h ( 2 ) = 6 h ( 2 ) = 6

The domain of function f − 1 f − 1 is ( − ∞ , − 2 ) ( − ∞ , − 2 ) and the range of function f − 1 f − 1 is ( 1 , ∞ ) . ( 1 , ∞ ) .

  • f ( 60 ) = 50. f ( 60 ) = 50. In 60 minutes, 50 miles are traveled.
  • f − 1 ( 60 ) = 70. f − 1 ( 60 ) = 70. To travel 60 miles, it will take 70 minutes.

a. 3; b. 5.6

x = 3 y + 5 x = 3 y + 5

f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ] f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ]

1.1 Section Exercises

A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.

When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.

When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.

not a function

f ( − 3 ) = − 11 ; f ( − 3 ) = − 11 ; f ( 2 ) = − 1 ; f ( 2 ) = − 1 ; f ( − a ) = − 2 a − 5 ; f ( − a ) = − 2 a − 5 ; − f ( a ) = − 2 a + 5 ; − f ( a ) = − 2 a + 5 ; f ( a + h ) = 2 a + 2 h − 5 f ( a + h ) = 2 a + 2 h − 5

f ( − 3 ) = 5 + 5 ; f ( − 3 ) = 5 + 5 ; f ( 2 ) = 5 ; f ( 2 ) = 5 ; f ( − a ) = 2 + a + 5 ; f ( − a ) = 2 + a + 5 ; − f ( a ) = − 2 − a − 5 ; − f ( a ) = − 2 − a − 5 ; f ( a + h ) = 2 − a − h + 5 f ( a + h ) = 2 − a − h + 5

f ( − 3 ) = 2 ; f ( − 3 ) = 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( − a ) = | − a − 1 | − | − a + 1 | ; f ( − a ) = | − a − 1 | − | − a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; f ( a + h ) = | a + h − 1 | − | a + h + 1 | f ( a + h ) = | a + h − 1 | − | a + h + 1 |

g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a

  • ⓐ f ( − 2 ) = 14 ; f ( − 2 ) = 14 ;
  • ⓑ x = 3 x = 3
  • ⓐ f ( 5 ) = 10 ; f ( 5 ) = 10 ;
  • ⓑ x = − 1 x = − 1 or x = 4 x = 4
  • ⓐ f ( t ) = 6 − 2 3 t ; f ( t ) = 6 − 2 3 t ;
  • ⓑ f ( − 3 ) = 8 ; f ( − 3 ) = 8 ;
  • ⓒ t = 6 t = 6
  • ⓐ f ( 0 ) = 1 ; f ( 0 ) = 1 ;
  • ⓑ f ( x ) = − 3 , x = − 2 f ( x ) = − 3 , x = − 2 or x = 2 x = 2

not a function so it is also not a one-to-one function

one-to-one function

function, but not one-to-one

f ( x ) = 1 , x = 2 f ( x ) = 1 , x = 2

f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2 f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2

f ( − 2 ) = 4 ;    f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236 f ( − 2 ) = 4 ;    f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236

f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9 f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9

[ 0 ,  100 ] [ 0 ,  100 ]

[ − 0.001 ,  0 .001 ] [ − 0.001 ,  0 .001 ]

[ − 1 , 000 , 000 ,  1,000,000 ] [ − 1 , 000 , 000 ,  1,000,000 ]

[ 0 ,  10 ] [ 0 ,  10 ]

[ −0.1 , 0.1 ] [ −0.1 , 0.1 ]

[ − 100 ,  100 ] [ − 100 ,  100 ]

  • ⓐ g ( 5000 ) = 50 ; g ( 5000 ) = 50 ;
  • ⓑ The number of cubic yards of dirt required for a garden of 100 square feet is 1.
  • ⓐ The height of a rocket above ground after 1 second is 200 ft.
  • ⓑ the height of a rocket above ground after 2 seconds is 350 ft.

1.2 Section Exercises

The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.

There is no restriction on x x for f ( x ) = x 3 f ( x ) = x 3 because you can take the cube root of any real number. So the domain is all real numbers, ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . When dealing with the set of real numbers, you cannot take the square root of negative numbers. So x x -values are restricted for f ( x ) = x f ( x ) = x to nonnegative numbers and the domain is [ 0 , ∞ ) . [ 0 , ∞ ) .

Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the x x -axis and y y -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate − ∞ − ∞ or ∞ . ∞ . Combine the graphs to find the graph of the piecewise function.

( − ∞ , 3 ] ( − ∞ , 3 ]

( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ ) ( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ )

( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ ) ( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ )

( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ ) ( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ )

( − ∞ , 5 ) ( − ∞ , 5 )

[ 6 , ∞ ) [ 6 , ∞ )

( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ ) ( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ )

domain: ( 2 , 8 ] , ( 2 , 8 ] , range [ 6 , 8 ) [ 6 , 8 )

domain: [ − 4 ,  4], [ − 4 ,  4], range: [ 0 ,  2] [ 0 ,  2]

domain: [ − 5 , 3 ) , [ − 5 , 3 ) , range: [ 0 , 2 ] [ 0 , 2 ]

domain: ( − ∞ , 1 ] , ( − ∞ , 1 ] , range: [ 0 , ∞ ) [ 0 , ∞ )

domain: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; range: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ]

domain: [ − 3 , ∞ ) ; [ − 3 , ∞ ) ; range: [ 0 , ∞ ) [ 0 , ∞ )

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0 f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0

f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34 f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34

f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16 f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16

domain: ( − ∞ , 1 ) ∪ ( 1 , ∞ ) ( − ∞ , 1 ) ∪ ( 1 , ∞ )

window: [ − 0.5 , − 0.1 ] ; [ − 0.5 , − 0.1 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

window: [ 0.1 , 0.5 ] ; [ 0.1 , 0.5 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

[ 0 , 8 ] [ 0 , 8 ]

Many answers. One function is f ( x ) = 1 x − 2 . f ( x ) = 1 x − 2 .

1.3 Section Exercises

Yes, the average rate of change of all linear functions is constant.

The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.

4 ( b + 1 ) 4 ( b + 1 )

4 x + 2 h 4 x + 2 h

− 1 13 ( 13 + h ) − 1 13 ( 13 + h )

3 h 2 + 9 h + 9 3 h 2 + 9 h + 9

4 x + 2 h − 3 4 x + 2 h − 3

increasing on ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , decreasing on ( − 2.5 , 1 ) ( − 2.5 , 1 )

increasing on ( − ∞ , 1 ) ∪ ( 3 , 4 ) , ( − ∞ , 1 ) ∪ ( 3 , 4 ) , decreasing on ( 1 , 3 ) ∪ ( 4 , ∞ ) ( 1 , 3 ) ∪ ( 4 , ∞ )

local maximum: ( − 3 , 50 ) , ( − 3 , 50 ) , local minimum: ( 3 , − 50 ) ( 3 , − 50 )

absolute maximum at approximately ( 7 , 150 ) , ( 7 , 150 ) , absolute minimum at approximately ( −7.5 , −220 ) ( −7.5 , −220 )

a. –3000; b. –1250

Local minimum at ( 3 , − 22 ) , ( 3 , − 22 ) , decreasing on ( − ∞ , 3 ) , ( − ∞ , 3 ) , increasing on ( 3 , ∞ ) ( 3 , ∞ )

Local minimum at ( − 2 , − 2 ) , ( − 2 , − 2 ) , decreasing on ( − 3 , − 2 ) , ( − 3 , − 2 ) , increasing on ( − 2 , ∞ ) ( − 2 , ∞ )

Local maximum at ( − 0.5 , 6 ) , ( − 0.5 , 6 ) , local minima at ( − 3.25 , − 47 ) ( − 3.25 , − 47 ) and ( 2.1 , − 32 ) , ( 2.1 , − 32 ) , decreasing on ( − ∞ , − 3.25 ) ( − ∞ , − 3.25 ) and ( − 0.5 , 2.1 ) , ( − 0.5 , 2.1 ) , increasing on ( − 3.25 , − 0.5 ) ( − 3.25 , − 0.5 ) and ( 2.1 , ∞ ) ( 2.1 , ∞ )

b = 5 b = 5

2.7 gallons per minute

approximately –0.6 milligrams per day

1.4 Section Exercises

Find the numbers that make the function in the denominator g g equal to zero, and check for any other domain restrictions on f f and g , g , such as an even-indexed root or zeros in the denominator.

Yes. Sample answer: Let f ( x ) = x + 1  and  g ( x ) = x − 1. f ( x ) = x + 1  and  g ( x ) = x − 1. Then f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x and g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . So f ∘ g = g ∘ f . f ∘ g = g ∘ f .

( f + g ) ( x ) = 2 x + 6 , ( f + g ) ( x ) = 2 x + 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f − g ) ( x ) = 2 x 2 + 2 x − 6 , ( f − g ) ( x ) = 2 x 2 + 2 x − 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , ( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = x 2 + 2 x 6 − x 2 , ( f g ) ( x ) = x 2 + 2 x 6 − x 2 , domain: ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ )

( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , ( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , ( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = x + 2 , ( f g ) ( x ) = x + 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = 4 x 3 + 8 x 2 , ( f g ) ( x ) = 4 x 3 + 8 x 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f + g ) ( x ) = 3 x 2 + x − 5 , ( f + g ) ( x ) = 3 x 2 + x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f − g ) ( x ) = 3 x 2 − x − 5 , ( f − g ) ( x ) = 3 x 2 − x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: ( 5 , ∞ ) ( 5 , ∞ )

  • ⓑ f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1 ; f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1 ;
  • ⓒ g ( f ) ( x ) ) = 6 x 2 − 2 ; g ( f ) ( x ) ) = 6 x 2 − 2 ;
  • ⓓ ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20 ; ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20 ;
  • ⓔ ( f ∘ f ) ( − 2 ) = 163 ( f ∘ f ) ( − 2 ) = 163

f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7 f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7

f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x

( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4 ( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4

f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1 f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1

  • ⓐ Text ( g ∘ f ) ( x ) = − 3 2 − 4 x ; ( g ∘ f ) ( x ) = − 3 2 − 4 x ;
  • ⓑ ( − ∞ , 1 2 ) ( − ∞ , 1 2 )
  • ⓐ ( 0 , 2 ) ∪ ( 2 , ∞ ) ; ( 0 , 2 ) ∪ ( 2 , ∞ ) ;
  • ⓑ ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; c. ( 0 , ∞ ) ( 0 , ∞ )

( 1 , ∞ ) ( 1 , ∞ )

sample: f ( x ) = x 3 g ( x ) = x − 5 f ( x ) = x 3 g ( x ) = x − 5

sample: f ( x ) = 4 x g ( x ) = ( x + 2 ) 2 f ( x ) = 4 x g ( x ) = ( x + 2 ) 2

sample: f ( x ) = x 3 g ( x ) = 1 2 x − 3 f ( x ) = x 3 g ( x ) = 1 2 x − 3

sample: f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5 f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5

sample: f ( x ) = x f ( x ) = x g ( x ) = 2 x + 6 g ( x ) = 2 x + 6

sample: f ( x ) = x 3 f ( x ) = x 3 g ( x ) = ( x − 1 ) g ( x ) = ( x − 1 )

sample: f ( x ) = x 3 f ( x ) = x 3 g ( x ) = 1 x − 2 g ( x ) = 1 x − 2

sample: f ( x ) = x f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 g ( x ) = 2 x − 1 3 x + 4

f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94 f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94

f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5 f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5

18 x 2 + 60 x + 51 18 x 2 + 60 x + 51

g ∘ g ( x ) = 9 x + 20 g ∘ g ( x ) = 9 x + 20

( f ∘ g ) ( 6 ) = 6 ( f ∘ g ) ( 6 ) = 6 ; ( g ∘ f ) ( 6 ) = 6 ( g ∘ f ) ( 6 ) = 6

( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11 ( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11

A ( t ) = π ( 25 t + 2 ) 2 A ( t ) = π ( 25 t + 2 ) 2 and A ( 2 ) = π ( 25 4 ) 2 = 2500 π A ( 2 ) = π ( 25 4 ) 2 = 2500 π square inches

A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π square units

  • ⓐ N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1 ; N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1 ;
  • ⓑ 3.38 hours

1.5 Section Exercises

A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.

For a function f , f , substitute ( − x ) ( − x ) for ( x ) ( x ) in f ( x ) . f ( x ) . Simplify. If the resulting function is the same as the original function, f ( − x ) = f ( x ) , f ( − x ) = f ( x ) , then the function is even. If the resulting function is the opposite of the original function, f ( − x ) = − f ( x ) , f ( − x ) = − f ( x ) , then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

g ( x ) = | x - 1 | − 3 g ( x ) = | x - 1 | − 3

g ( x ) = 1 ( x + 4 ) 2 + 2 g ( x ) = 1 ( x + 4 ) 2 + 2

The graph of f ( x + 43 ) f ( x + 43 ) is a horizontal shift to the left 43 units of the graph of f . f .

The graph of f ( x - 4 ) f ( x - 4 ) is a horizontal shift to the right 4 units of the graph of f . f .

The graph of f ( x ) + 8 f ( x ) + 8 is a vertical shift up 8 units of the graph of f . f .

The graph of f ( x ) − 7 f ( x ) − 7 is a vertical shift down 7 units of the graph of f . f .

The graph of f ( x + 4 ) − 1 f ( x + 4 ) − 1 is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f . f .

decreasing on ( − ∞ , − 3 ) ( − ∞ , − 3 ) and increasing on ( − 3 , ∞ ) ( − 3 , ∞ )

decreasing on [ 0 , ∞ ) [ 0 , ∞ )

g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1 g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1

f ( x ) = | x - 3 | − 2 f ( x ) = | x - 3 | − 2

f ( x ) = x + 3 − 1 f ( x ) = x + 3 − 1

f ( x ) = ( x - 2 ) 2 f ( x ) = ( x - 2 ) 2

f ( x ) = | x + 3 | − 2 f ( x ) = | x + 3 | − 2

f ( x ) = − x f ( x ) = − x

f ( x ) = − ( x + 1 ) 2 + 2 f ( x ) = − ( x + 1 ) 2 + 2

f ( x ) = − x + 1 f ( x ) = − x + 1

The graph of g g is a vertical reflection (across the x x -axis) of the graph of f . f .

The graph of g g is a vertical stretch by a factor of 4 of the graph of f . f .

The graph of g g is a horizontal compression by a factor of 1 5 1 5 of the graph of f . f .

The graph of g g is a horizontal stretch by a factor of 3 of the graph of f . f .

The graph of g g is a horizontal reflection across the y y -axis and a vertical stretch by a factor of 3 of the graph of f . f .

g ( x ) = | − 4 x | g ( x ) = | − 4 x |

g ( x ) = 1 3 ( x + 2 ) 2 − 3 g ( x ) = 1 3 ( x + 2 ) 2 − 3

g ( x ) = 1 2 ( x - 5 ) 2 + 1 g ( x ) = 1 2 ( x - 5 ) 2 + 1

The graph of the function f ( x ) = x 2 f ( x ) = x 2 is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.

The graph of f ( x ) = | x | f ( x ) = | x | is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.

The graph of the function f ( x ) = x 3 f ( x ) = x 3 is compressed vertically by a factor of 1 2 . 1 2 .

The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.

The graph of f ( x ) = x f ( x ) = x is shifted right 4 units and then reflected across the vertical line x = 4. x = 4.

1.6 Section Exercises

Isolate the absolute value term so that the equation is of the form | A | = B . | A | = B . Form one equation by setting the expression inside the absolute value symbol, A , A , equal to the expression on the other side of the equation, B . B . Form a second equation by setting A A equal to the opposite of the expression on the other side of the equation, − B . − B . Solve each equation for the variable.

The graph of the absolute value function does not cross the x x -axis, so the graph is either completely above or completely below the x x -axis.

First determine the boundary points by finding the solution(s) of the equation. Use the boundary points to form possible solution intervals. Choose a test value in each interval to determine which values satisfy the inequality.

| x + 4 | = 1 2 | x + 4 | = 1 2

| f ( x ) − 8 | < 0.03 | f ( x ) − 8 | < 0.03

{ 1 , 11 } { 1 , 11 }

{ - 9 4 , 13 4 } { - 9 4 , 13 4 }

{ 10 3 , 20 3 } { 10 3 , 20 3 }

{ 11 5 , 29 5 } { 11 5 , 29 5 }

{ 5 2 , 7 2 } { 5 2 , 7 2 }

No solution

{ − 57 , 27 } { − 57 , 27 }

( 0 , − 8 ) ; ( − 6 , 0 ) , ( 4 , 0 ) ( 0 , − 8 ) ; ( − 6 , 0 ) , ( 4 , 0 )

( 0 , − 7 ) ; ( 0 , − 7 ) ; no x x -intercepts

( − ∞ , − 8 ) ∪ ( 12 , ∞ ) ( − ∞ , − 8 ) ∪ ( 12 , ∞ )

− 4 3 ≤ x ≤ 4 − 4 3 ≤ x ≤ 4

( − ∞ , − 8 3 ] ∪ [ 6 , ∞ ) ( − ∞ , − 8 3 ] ∪ [ 6 , ∞ )

( − ∞ , − 8 3 ] ∪ [ 16 , ∞ ) ( − ∞ , − 8 3 ] ∪ [ 16 , ∞ )

range: [ 0 , 20 ] [ 0 , 20 ]

x - x - intercepts:

There is no solution for a a that will keep the function from having a y y -intercept. The absolute value function always crosses the y y -intercept when x = 0. x = 0.

| p − 0.08 | ≤ 0.015 | p − 0.08 | ≤ 0.015

| x − 5.0 | ≤ 0.01 | x − 5.0 | ≤ 0.01

1.7 Section Exercises

Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that y y -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no y y -values repeat and the function is one-to-one.

Yes. For example, f ( x ) = 1 x f ( x ) = 1 x is its own inverse.

Given a function y = f ( x ) , y = f ( x ) , solve for x x in terms of y . y . Interchange the x x and y . y . Solve the new equation for y . y . The expression for y y is the inverse, y = f − 1 ( x ) . y = f − 1 ( x ) .

f − 1 ( x ) = x − 3 f − 1 ( x ) = x − 3

f − 1 ( x ) = 2 − x f − 1 ( x ) = 2 − x

f − 1 ( x ) = − 2 x x − 1 f − 1 ( x ) = − 2 x x − 1

domain of f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7 f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7

domain of f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5 f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5

  • ⓐ f ( g ( x ) ) = x f ( g ( x ) ) = x and g ( f ( x ) ) = x . g ( f ( x ) ) = x .
  • ⓑ This tells us that f f and g g are inverse functions

f ( g ( x ) ) = x , g ( f ( x ) ) = x f ( g ( x ) ) = x , g ( f ( x ) ) = x

not one-to-one

[ 2 , 10 ] [ 2 , 10 ]

f − 1 ( x ) = ( 1 + x ) 1 / 3 f − 1 ( x ) = ( 1 + x ) 1 / 3

f − 1 ( x ) = 5 9 ( x − 32 ) . f − 1 ( x ) = 5 9 ( x − 32 ) . Given the Fahrenheit temperature, x , x , this formula allows you to calculate the Celsius temperature.

t ( d ) = d 50 , t ( d ) = d 50 , t ( 180 ) = 180 50 . t ( 180 ) = 180 50 . The time for the car to travel 180 miles is 3.6 hours.

f ( − 3 ) = − 27 ; f ( − 3 ) = − 27 ; f ( 2 ) = − 2 ; f ( 2 ) = − 2 ; f ( − a ) = − 2 a 2 − 3 a ; f ( − a ) = − 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2 f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2

x = − 1.8 x = − 1.8 or  or  x = 1.8  or  x = 1.8

− 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64 − 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64

( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ )

increasing ( 2 , ∞ ) ; ( 2 , ∞ ) ; decreasing ( − ∞ , 2 ) ( − ∞ , 2 )

increasing ( − 3 , 1 ) ; ( − 3 , 1 ) ; constant ( − ∞ , − 3 ) ∪ ( 1 , ∞ ) ( − ∞ , − 3 ) ∪ ( 1 , ∞ )

local minimum ( − 2 , − 3 ) ; ( − 2 , − 3 ) ; local maximum ( 1 , 3 ) ( 1 , 3 )

Absolute Maximum: 10

( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x ( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x

( f ∘ g ) ( x ) = 1 x + 2 ; ( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 x + 2 ( g ∘ f ) ( x ) = 1 x + 2

( f ∘ g ) ( x ) = 1 + x 1 + 4 x , x ≠ 0 , x ≠ − 1 4 ( f ∘ g ) ( x ) = 1 + x 1 + 4 x , x ≠ 0 , x ≠ − 1 4

( f ∘ g ) ( x ) = 1 x , x > 0 ( f ∘ g ) ( x ) = 1 x , x > 0

sample: g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x

f ( x ) = | x − 3 | f ( x ) = | x − 3 |

f ( x ) = 1 2 | x + 2 | + 1 f ( x ) = 1 2 | x + 2 | + 1

f ( x ) = − 3 | x − 3 | + 3 f ( x ) = − 3 | x − 3 | + 3

x = − 22 , x = 14 x = − 22 , x = 14

( − 5 3 , 3 ) ( − 5 3 , 3 )

f − 1 ( x ) = x - 1 f − 1 ( x ) = x - 1

The function is one-to-one.

The function is not one-to-one.

The relation is a function.

The graph is a parabola and the graph fails the horizontal line test.

2 a 2 − a 2 a 2 − a

− 2 ( a + b ) + 1 − 2 ( a + b ) + 1

x = − 7 x = − 7 and x = 10 x = 10

f − 1 ( x ) = x + 5 3 f − 1 ( x ) = x + 5 3

( − ∞ , − 1.1 )  and  ( 1.1 , ∞ ) ( − ∞ , − 1.1 )  and  ( 1.1 , ∞ )

( 1.1 , − 0.9 ) ( 1.1 , − 0.9 )

f ( 2 ) = 2 f ( 2 ) = 2

f ( x ) = { | x | if x ≤ 2 3 if x > 2 f ( x ) = { | x | if x ≤ 2 3 if x > 2

x = 2 x = 2

f − 1 ( x ) = − x − 11 2 f − 1 ( x ) = − x − 11 2

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Inequalities Unit 7th Grade CCSS includes solving one and two-step inequalities, writing inequalities, and solving inequalities in real-world situations. | maneuveringthemiddle.com

Inequalities Unit 7th Grade CCSS

A 7-day CCSS-Aligned Inequalities Unit which includes solving one-step inequalities, solving two-step inequalities, writing inequalities, and solving inequalities in real-world situations.

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Inequalities Unit 7th Grade CCSS – A 7 day CCSS-Aligned Inequalities Unit which includes solving one-step inequalities, solving two-step inequalities, writing inequalities, and solving inequalities in real world situations.

Students will practice with both skill-based problems, real-world application questions, and error analysis to support higher level thinking skills.  You can reach your students and teach the standards without all of the prep and stress of creating materials!

Standards:   7.EE.4 and 7.EE.4B;  Texas Teacher?  Grab the TEKS-Aligned Equations and Inequalities Unit.  Please don’t purchase both as there is overlapping content.

Learning Focus:

  • solve two-step inequalities and represent the solution on a number line
  • interpret the solution of an inequality
  • write and solve two-step inequalities that represent a real-world situation

What is included in the 7th grade ccss Inequalities Unit?

1. Unit Overviews

  • Streamline planning with unit overviews that include essential questions, big ideas, vertical alignment, vocabulary, and common misconceptions.
  • A pacing guide and tips for teaching each topic are included to help you be more efficient in your planning.

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  • Available as a PDF and the student handouts/homework/study guides have been converted to Google Slides™ for your convenience.

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  • Daily homework is aligned directly to the student handouts and is versatile for both in class or at home practice.

4. Assessments

  • 1-2 quizzes, a unit study guide, and a unit test allow you to easily assess and meet the needs of your students.
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***Please download a preview to see sample pages and more information.***

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  • Incorporate our  Inequalities Activity Bundle  for hands-on activities as additional and engaging practice opportunities.

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  • Each student handout is designed for a single class period. However, feel free to review the problems and select specific ones to meet your student needs. There are multiple problems to practice the same concepts, so you can adjust as needed.

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  • The unit test is editable with Microsoft PPT. The remainder of the file is a PDF and not editable.

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These guided notes and assignments have been a lifesaver and 100% worth the cost.. and I’m on a teacher’s salary so I tend to be very cheap! I am a new teacher and these resources are well organized, easy to follow, student-friendly, and standards-based. If you’re going to spend money on something for your classroom, I recommend buying this.

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Unit 5: Equations & inequalities introduction

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Common Core Algebra I Math (Worksheets, Homework, Lesson Plans)

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Introduction to Inequalities Homework

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This is the accompanying homework for the Introduction to Inequalities notes page (the freebie on my store page!). It includes representing inequalities graphically (on a number line), translating a verbal inequality into an inequality with symbols and graphing it, and stating whether a given number is in the solution set of an inequality. An answer key is included.

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1.7: Solve Absolute Value Inequalities

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Learning Objectives

By the end of this section, you will be able to:

  • Solve absolute value equations
  • Solve absolute value inequalities with “less than”
  • Solve absolute value inequalities with “greater than”
  • Solve applications with absolute value

Before you get started, take this readiness quiz.

  • Evaluate: \(−|7|\). If you missed this problem, review [link] .
  • Fill in \(<,>,<,>,\) or \(=\) for each of the following pairs of numbers. ⓐ \(|−8|\text{___}−|−8|\) ⓑ \(12\text{___}−|−12|\) ⓒ \(|−6|\text{___}−6\) ⓓ \(−(−15)\text{___}−|−15|\) If you missed this problem, review [link] .
  • Simplify: \(14−2|8−3(4−1)|\). If you missed this problem, review [link] .

Solve Absolute Value Equations

As we prepare to solve absolute value equations, we review our definition of absolute value .

ABSOLUTE VALUE

The absolute value of a number is its distance from zero on the number line.

The absolute value of a number n is written as \(|n|\) and \(|n|\geq 0\) for all numbers.

Absolute values are always greater than or equal to zero.

We learned that both a number and its opposite are the same distance from zero on the number line. Since they have the same distance from zero, they have the same absolute value. For example:

  • \(−5\) is 5 units away from 0, so \(|−5|=5\).
  • \(5\) is 5 units away from 0, so \(|5|=5\).

Figure \(\PageIndex{1}\) illustrates this idea.

The figure is a number line with tick marks at negative 5, 0, and 5. The distance between negative 5 and 0 is given as 5 units, so the absolute value of negative 5 is 5. The distance between 5 and 0 is 5 units, so the absolute value of 5 is 5.

For the equation |x|=5,|x|=5, we are looking for all numbers that make this a true statement. We are looking for the numbers whose distance from zero is 5. We just saw that both 5 and −5−5 are five units from zero on the number line. They are the solutions to the equation.

\(\begin{array} {ll} {\text{If}} &{|x|=5} \\ {\text{then}} &{x=−5\text{ or }x=5} \\ \end{array}\)

The solution can be simplified to a single statement by writing \(x=\pm 5\). This is read, “ x is equal to positive or negative 5”.

We can generalize this to the following property for absolute value equations.

ABSOLUTE VALUE EQUATIONS

For any algebraic expression, u , and any positive real number, a ,

\[\begin{array} {ll} {\text{if}} &{|u|=a} \\ {\text{then}} &{u=−a \text{ or }u=a} \\ \nonumber \end{array}\]

Remember that an absolute value cannot be a negative number.

Example \(\PageIndex{1}\)

  • \(|y|=−6\)

\(\begin{array} {ll} {} &{|x|=8} \\ {\text{Write the equivalent equations.}} &{x=−8 \text{ or } x=8} \\ {} &{x=\pm 8} \\ \end{array}\)

\(\begin{array} {ll} {} &{|y|=−6} \\ {} &{\text{No solution}} \\ \end{array}\) Since an absolute value is always positive, there are no solutions to this equation.

\(\begin{array} {ll} {} &{|z|=0} \\ {\text{Write the equivalent equations.}} &{z=−0\text{ or }z=0} \\ {\text{Since }−0=0,} &{z=0} \\ \end{array}\) Both equations tell us that z=0z=0 and so there is only one solution.

EXERCISE \(\PageIndex{2}\)

  • \(|y|=−4\)

no solution

EXERCISE \(\PageIndex{3}\)

  • \(|y|=−5\)

To solve an absolute value equation , we first isolate the absolute value expression using the same procedures we used to solve linear equations. Once we isolate the absolute value expression we rewrite it as the two equivalent equations.

How to Solve Absolute Value Equations

Example \(\pageindex{4}\).

Solve \(|5x−4|−3=8\).

Step 1 is to isolate the absolute value expression. The difference between the absolute value of the quantity 5 x minus 4 and 3 is equal to 8. Add 3 to both sides. The result is the absolute value of the quantity 5 x minus 4 is equal to 11.

EXERCISE \(\PageIndex{5}\)

Solve: \(|3x−5|−1=6\).

\(x=4, \space x=−\frac{2}{3}\)

EXERCISE \(\PageIndex{6}\)

Solve: \(|4x−3|−5=2\).

\(x=−1,\space x=\frac{5}{2}\)

The steps for solving an absolute value equation are summarized here.

SOLVE ABSOLUTE VALUE EQUATIONS.

  • Isolate the absolute value expression.
  • Write the equivalent equations.
  • Solve each equation.
  • Check each solution.

Example \(\PageIndex{7}\)

Solve \(2|x−7|+5=9\).

Exercise \(\PageIndex{8}\)

Solve: \(3|x−4|−4=8\).

\(x=8,\space x=0\)

Exercise \(\PageIndex{9}\)

Solve: \(2|x−5|+3=9\).

\(x=8,\space x=2\)

Remember, an absolute value is always positive!

Example \(\PageIndex{10}\)

Solve: \(|\frac{2}{3}x−4|+11=3\).

\(\begin{array} {ll} {} &{|\frac{2}{3}x−4|=−8} \\ {\text{Isolate the absolute value term.}} &{|\frac{2}{3}x−4|=−8} \\ {\text{An absolute value cannot be negative.}} &{\text{No solution}} \\ \end{array}\)

Exercise \(\PageIndex{11}\)

Solve: \(|\frac{3}{4}x−5|+9=4\).

No solution

Exercise \(\PageIndex{12}\)

Solve: \(|\frac{5}{6}x+3|+8=6\).

Some of our absolute value equations could be of the form \(|u|=|v|\) where u and v are algebraic expressions. For example, \(|x−3|=|2x+1|\).

How would we solve them? If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. The property for absolute value equations says that for any algebraic expression, u , and a positive real number, a , if \(|u|=a\), then \(u=−a\) or \(u=a\).

This tell us that

\(\begin{array} {llll} {\text{if}} &{|u|=|v|} &{} &{} \\ {\text{then}} &{|u|=v} &{\text{or}} &{|u|=−v} \\ {\text{and so}} &{u=v \text{ or } u = −v} &{\text{or}} &{u=−v \text{ or } u = −(−v)} \\ \end{array}\)

This leads us to the following property for equations with two absolute values.

EQUATIONS WITH TWO ABSOLUTE VALUES

For any algebraic expressions, u and v ,

\[\begin{array} {ll} {\text{if}} &{|u|=|v|} \\ {\text{then}} &{u=−v\text{ or }u=v} \\ \nonumber \end{array}\]

When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed.

Example \(\PageIndex{13}\)

Solve: \(|5x−1|=|2x+3|\).

\(\begin{array} {ll} {} &{} &{|5x−1|=|2x+3|} &{} \\ {} &{} &{} &{} \\ {\text{Write the equivalent equations.}} &{5x−1=−(2x+3)} &{\text{or}} &{5x−1=2x+3} \\ {} &{5x−1=−2x−3} &{\text{or}} &{3x−1=3} \\ {\text{Solve each equation.}} &{7x−1=−3} &{} &{3x=4} \\ {} &{7x=−2} &{} &{x=43} \\ {} &{x=−27} &{\text{or}} &{x=43} \\ {\text{Check.}} &{} &{} &{} \\ {\text{We leave the check to you.}} &{} &{} &{} \\ \end{array}\)

Exercise \(\PageIndex{14}\)

Solve: \(|7x−3|=|3x+7|\).

\(x=−\frac{2}{5}, \space x=\frac{5}{2}\)

Exercise \(\PageIndex{15}\)

Solve: \(|6x−5|=|3x+4|\).

\(x=3, x=19\)

Solve Absolute Value Inequalities with “Less Than”

Let’s look now at what happens when we have an absolute value inequality . Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work. Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line. For the equation \(|x|=5\), we saw that both 5 and \(−5\) are five units from zero on the number line. They are the solutions to the equation.

\[\begin{array} {lll} {} &{|x|=5} &{} \\ {x=−5} &{\text{or}} &{x=5} \\ \nonumber \end{array}\]

What about the inequality \(|x|\leq 5\)? Where are the numbers whose distance is less than or equal to 5? We know \(−5\) and 5 are both five units from zero. All the numbers between \(−5\) and 5 are less than five units from zero (Figure \(\PageIndex{2}\)).

The figure is a number line with negative 5, 0, and 5 displayed. There is a left bracket at negative 5 and a right bracket at 5. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x which is less than or equal to 5.

In a more general way, we can see that if \(|u|\leq a\), then \(−a\leq u\leq a\) (Figure \(\PageIndex{3}\)).

The figure is a number line with negative a 0, and a displayed. There is a left bracket at negative a and a right bracket at a. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is less than or equal to a, then negative a is less than or equal to u which is less than or equal to a.

This result is summarized here.

ABSOLUTE VALUE INEQUALITIES WITH \(<\) OR \(\leq\)

\[ \text{if} \quad |u|<a, \quad \text{then} \space −a<u<a \\ \text{if} \quad |u|\leq a, \quad \text{then} \space−a\leq u\leq a \nonumber\]

After solving an inequality, it is often helpful to check some points to see if the solution makes sense. The graph of the solution divides the number line into three sections. Choose a value in each section and substitute it in the original inequality to see if it makes the inequality true or not. While this is not a complete check, it often helps verify the solution.

Example \(\PageIndex{16}\)

Solve \(|x|<7\). Graph the solution and write the solution in interval notation.

To verify, check a value in each section of the number line showing the solution. Choose numbers such as −8,−8, 1, and 9.

The figure is a number line with a left parenthesis at negative 7, a right parenthesis at 7 and shading between the parentheses. The values negative 8, 1, and 9 are marked with points. The absolute value of negative 8 is less than 7 is false. It does not satisfy the absolute value of x is less than 7. The absolute value of 1 is less than 7 is true. It does satisfy the absolute value of x is less than 7. The absolute value of 9 is less than 7 is false. It does not satisfy the absolute value of x is less than 7.

EXERCISE \(\PageIndex{17}\)

Graph the solution and write the solution in interval notation: \(|x|<9\).

The solution is negative 9 is less than x which is less than 9. The number line shows open circles at negative 9 and 9 with shading in between the circles. The interval notation is negative 9 to 9 within parentheses.

EXERCISE \(\PageIndex{18}\)

Graph the solution and write the solution in interval notation: \(|x|<1\).

The solution is negative 1 is less than x which is less than 1. The number line shows open circles at negative 1 and 1 with shading in between the circles. The interval notation is negative 1 to 1 within parentheses.

Example \(\PageIndex{19}\)

Solve \(|5x−6|\leq 4\). Graph the solution and write the solution in interval notation.

EXERCISE \(\PageIndex{20}\)

Solve \(|2x−1|\leq 5\). Graph the solution and write the solution in interval notation:

The solution is negative 2 is less than or equal to x which is less than or equal to 3. The number line shows closed circles at negative 2 and 3 with shading between the circles. The interval notation is negative 2 to 3 within brackets.

EXERCISE \(\PageIndex{21}\)

Solve \(|4x−5|\leq 3\). Graph the solution and write the solution in interval notation:

The solution is one-half is less than or equal to x which is less than or equal to 2. The number line shows closed circles at one-half and 2 with shading between the circles. The interval notation is one-half to 2 within brackets.

SOLVE ABSOLUTE VALUE INEQUALITIES WITH \(<\) OR \(\leq\)

\[\begin{array} {lll} {|u|<a} &{\quad \text{is equivalent to}} &{−a<u<a} \\ {|u|\leq a} &{\quad \text{is equivalent to}} &{−a\leq u\leq a} \\ \nonumber \end{array}\]

  • Solve the compound inequality.
  • Graph the solution
  • Write the solution using interval notation.

Solve Absolute Value Inequalities with “Greater Than”

What happens for absolute value inequalities that have “greater than”? Again we will look at our definition of absolute value. The absolute value of a number is its distance from zero on the number line.

We started with the inequality \(|x|\leq 5\). We saw that the numbers whose distance is less than or equal to five from zero on the number line were \(−5\) and 5 and all the numbers between \(−5\) and 5 (Figure \(\PageIndex{4}\)).

The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its right and a right bracket at 5 with shading to its left. It illustrates that if the absolute value of x is less than or equal to 5, then negative 5 is less than or equal to x is less than or equal to 5.

Now we want to look at the inequality \(|x|\geq 5\). Where are the numbers whose distance from zero is greater than or equal to five?

Again both \(−5\) and 5 are five units from zero and so are included in the solution. Numbers whose distance from zero is greater than five units would be less than \(−5\) and greater than 5 on the number line (Figure \(\PageIndex{5}\)).

The figure is a number line with negative 5, 0, and 5 displayed. There is a right bracket at negative 5 that has shading to its left and a left bracket at 5 with shading to its right. The distance between negative 5 and 0 is given as 5 units and the distance between 5 and 0 is given as 5 units. It illustrates that if the absolute value of x is greater than or equal to 5, then x is less than or equal to negative 5 or x is greater than or equal to 5.

In a more general way, we can see that if \(|u|\geq a\), then \(u\leq −a\) or \(u\leq a\). See Figure .

The figure is a number line with negative a, 0, and a displayed. There is a right bracket at negative a that has shading to its left and a left bracket at a with shading to its right. The distance between negative a and 0 is given as a units and the distance between a and 0 is given as a units. It illustrates that if the absolute value of u is greater than or equal to a, then u is less than or equal to negative a or u is greater than or equal to a.

ABSOLUTE VALUE INEQUALITIES WITH \(>\) OR \(\geq\)

\[\begin{array} {lll} {\text{if}} &{\quad |u|>a,} &{\quad \text{then } u<−a \text{ or } u>a} \\ {\text{if}} &{\quad |u|\geq a,} &{\quad \text{then } u\leq −a \text{ or } u\geq a} \\ \nonumber \end{array}\]

Example \(\PageIndex{22}\)

Solve \(|x|>4\). Graph the solution and write the solution in interval notation.

To verify, check a value in each section of the number line showing the solution. Choose numbers such as −6,−6, 0, and 7.

The figure is a number line with a right parenthesis at negative 4 with shading to its left and a left parenthesis at 4 shading to its right. The values negative 6, 0, and 7 are marked with points. The absolute value of negative 6 is greater than negative 4 is true. It does not satisfy the absolute value of x is greater than 4. The absolute value of 0 is greater than 4 is false. It does not satisfy the absolute value of x is greater than 4. The absolute value of 7 is less than 4 is true. It does satisfy the absolute value of x is greater than 4.

EXERCISE \(\PageIndex{23}\)

Solve \(|x|>2\). Graph the solution and write the solution in interval notation.

The solution is x is less than negative 2 or x is greater than 2. The number line shows an open circle at negative 2 with shading to its left and an open circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative 2 within parentheses and 2 to infinity within parentheses.

EXERCISE \(\PageIndex{24}\)

Solve \(|x|>1\). Graph the solution and write the solution in interval notation.

The solution is x is less than negative 1 or x is greater than 1. The number line shows an open circle at negative 1 with shading to its left and an open circle at 1 with shading to its right. The interval notation is the union of negative infinity to negative 1 within parentheses and 1 to infinity within parentheses.

Example \(\PageIndex{25}\)

Solve \(|2x−3|\geq 5\). Graph the solution and write the solution in interval notation.

EXERCISE \(\PageIndex{26}\)

Solve \(|4x−3|\geq 5\). Graph the solution and write the solution in interval notation.

The solution is x is less than or equal to negative one-half or x is greater than or equal 2. The number line shows a closed circle at negative one-half with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to negative one-half within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis

EXERCISE \(\PageIndex{27}\)

Solve \(|3x−4|\geq 2\). Graph the solution and write the solution in interval notation.

The solution is x is less than or equal to two-thirds or x is greater than or equal 2. The number line shows a closed circle at two-thirds with shading to its left and a closed circle at 2 with shading to its right. The interval notation is the union of negative infinity to two-thirds within a parenthesis and a bracket and 2 to infinity within a bracket and a parenthesis.

SOLVE ABSOLUTE VALUE INEQUALITIES WITH \(>\) OR \(\geq\).

\[\begin{array} {lll} { |u| >a } &{\quad \text{is equivalent to}} &{ u<−a \quad \text{ or } \quad u>a} \\ { |u| \geq a } &{\quad \text{is equivalent to}} &{ u\leq −a \quad \text{ or } \quad u\geq a} \\ { |u| >a } &{\quad \text{is equivalent to}} &{ u<−a \quad \text{ or } \quad u>a} \\ { |u| \geq a } &{\quad \text{is equivalent to}} &{ u\leq −a \quad \text{ or } \quad u\geq a} \\ \nonumber \end{array}\]

Solve Applications with Absolute Value

Absolute value inequalities are often used in the manufacturing process. An item must be made with near perfect specifications. Usually there is a certain tolerance of the difference from the specifications that is allowed. If the difference from the specifications exceeds the tolerance, the item is rejected.

\[|\text{actual-ideal}|\leq \text{tolerance} \nonumber\]

Example \(\PageIndex{28}\)

The ideal diameter of a rod needed for a machine is 60 mm. The actual diameter can vary from the ideal diameter by \(0.075\) mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

\(\begin{array} {ll} {} &{\text{Let }x=\text{ the actual measurement}} \\ {\text{Use an absolute value inequality to express this situation.}} &{|\text{actual-ideal}|\leq \text{tolerance}} \\ {} &{|x−60|\leq 0.075} \\ {\text{Rewrite as a compound inequality.}} &{−0.075\leq x−60\leq 0.075} \\ {\text{Solve the inequality.}} &{59.925\leq x\leq 60.075} \\ {\text{Answer the question.}} &{\text{The diameter of the rod can be between}} \\ {} &{59.925 mm \text{ and } 60.075 mm.} \\ \end{array}\)

ExERCISE \(\PageIndex{29}\)

The ideal diameter of a rod needed for a machine is 80 mm. The actual diameter can vary from the ideal diameter by 0.009 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

The diameter of the rod can be between 79.991 and 80.009 mm.

ExERCISE \(\PageIndex{30}\)

The ideal diameter of a rod needed for a machine is 75 mm. The actual diameter can vary from the ideal diameter by 0.05 mm. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

The diameter of the rod can be between 74.95 and 75.05 mm.

Access this online resource for additional instruction and practice with solving linear absolute value equations and inequalities.

  • Solving Linear Absolute Value Equations and Inequalities

Key Concepts

  • Absolute Value The absolute value of a number is its distance from 0 on the number line. The absolute value of a number n is written as \(|n|\) and \(|n|\geq 0\) for all numbers. Absolute values are always greater than or equal to zero.
  • Absolute Value Equations For any algebraic expression, u , and any positive real number, a , \(\begin{array} {ll} {\text{if}} &{\quad |u|=a} \\ {\text{then}} &{\quad u=−a \text{ or } u=a} \\ \end{array}\) Remember that an absolute value cannot be a negative number.
  • Equations with Two Absolute Values For any algebraic expressions, u and v , \(\begin{array} {ll} {\text{if}} &{\quad |u|=|v|} \\ {\text{then}} &{\quad u=−v \text{ or } u=v} \\ \end{array}\)
  • Absolute Value Inequalities with \(<\) or \(\leq\) For any algebraic expression, u , and any positive real number, a , \(\begin{array} {llll} {\text{if}} &{\quad |u|=a} &{\quad \text{then}} &{−a<u<a} \\ {\text{if}} &{\quad |u|\leq a} &{\quad \text{then}} &{−a\leq u\leq a} \\ \end{array}\)
  • Write the equivalent compound inequality. \(\begin{array} {lll} {|u|<a} &{\quad \text{is equivalent to}} &{\quad −a<u<a} \\ {|u|\leq a} &{\quad \text{is equivalent to}} &{\quad −a\leq u\leq a} \\ \end{array}\)
  • Write the solution using interval notation
  • Absolute Value Inequalities with \(>\) or \(\geq\) For any algebraic expression, u , and any positive real number, a , \(\begin{array} {lll} {\text{if}} &{\quad |u|>a,} &{\text{then } u<−a\text{ or }u>a} \\ {\text{if}} &{\quad |u|\geq a,} &{\text{then } u\leq −a\text{ or }u\geq a} \\ \end{array}\)
  • Write the equivalent compound inequality. \(\begin{array} {lll} {|u|>a} &{\quad \text{is equivalent to}} &{\quad u<−a\text{ or }u>a} \\ {|u|\geq a} &{\quad \text{is equivalent to}} &{\quad u\leq −a\text{ or }u\geq a} \\ \end{array}\)

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  1. PDF Thank You for Your Purchase!

    19 Intro to Inequalities Homework 1 ... Additionally, an answer key is included. If you notice any discrepancies in the documents or have any questions, please email me at: [email protected]. ... Use your understanding of one-step inequalities to answer the questions below. 0 5 10 15 20 0 5 10 15 20 0 5 10 15 20 0 5 10 15 20.

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  8. Answer Key Chapter 1

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  10. Inequalities Unit 7th Grade CCSS

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    Student Handout 1 Homework 1 Student Handout 2 Homework 2 Student Handout 3 Homework 3 Quiz I Student Handout 4 ... Use your understanding of inequalities to answer the questions below. 4. 5. 6. C. Which inequality .s true when x _ 4? ... DAY 1 Intro to Inequalities Student Handout 1 Homework 1 DAY 6 Application of Inequalities Student Handout 5

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