2.1 Increasing, Decreasing, and Piecewise Functions; Applications

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2.1

Increasing, Decreasing, and Piecewise Functions; Applications

 Graph functions, looking for intervals on which the function is increasing, decreasing, or

constant, and estimate relative maxima and minima.

 Given an application, find a function that models the application; find the domain of the function and

function values, and then graph the function.

 Graph functions defined piecewise.

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Slide 2.1-2

Increasing, Decreasing, and Constant Functions

On a given interval, if the graph of a function rises from left to right, it is said to be increasing on that

interval. If the graph drops from left to right, it is said to be decreasing.

If the function values stay the same from left to right, the function is said to be constant.

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Definitions

A function f is said to be

increasing on an open interval I, if for all a and b in that

interval,

a < b implies f(a) < f(b).

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Slide 2.1-4

Definitions continued

A function f is said to be decreasing on an open

interval I, if for all a and b in that interval,

a < b implies f(a) > f(b).

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A function f is said to be

constant on an open interval I, if for all a and b in that

interval, f(a) = f(b).

Definitions continued

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Slide 2.1-6

Relative Maximum and Minimum Values

Suppose that f is a function for which f(c) exists for some c in the domain of f. Then:

f(c) is a relative maximum if there exists an open interval I containing c such that f(c) > f(x), for all x in I where x  c; and

f(c) is a relative minimum if there exists an open interval I containing c such that f(c) < f(x), for all x in I where x  c.

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Relative Maximum and Minimum Values

y

x c₁ c₂ c₃

Relative minimum Relative

maximum ^f

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Slide 2.1-8

Applications of Functions

Many real-world situations can be modeled by functions.

Example

A man plans to enclose a rectangular area using 80 yards of fencing. If the area is w yards wide, express the enclosed area as a function of w.

Solution

We want area as a function of w. Since the area is rectangular, we have A = lw.

We know that the perimeter, 2 lengths and 2 widths, is 80 yds, so we have 40 yds for one length and one width. If the width is w, then the length, l, can be given by l = 40 – w.

Now A(w) = (40 – w)w = 40w – w².

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Functions Defined Piecewise

For the function defined as:

find f (-3), f (1), and f (5).

2, for 0,

( ) 4, for 0 2, 1, for 2,

x x

f x x

x x

 

   

  



Some functions are defined piecewise using

different output formulas for different parts of the domain.

Since –3 0, use f (x) = x²: f (–3) = (–3)² = 9.

Since 0 < 1 2, use f (x) = 4: f (1) = 4.

Since 5 > 2 use f (x) = x – 1: f (5) = 5 – 1 = 4.



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Slide 2.1-10

c) We graph f(x) = only for inputs x greater than 2.

Functions Defined Piecewise

Graph the function defined as:

a) We graph f(x) = 3 only for inputs x less than or equal to 0.

b) We graph f(x) = 3 + x² only for inputs x greater than 0 and less than or equal to 2.

2

3 for 0

( ) 3 for 0 2

1 for 2 2

x

f x x x

x x

 



    

  



f(x) = 3, for x0

f(x) = 3 + x², for 0 < x2 ( ) 1for 2

2

f x  x x

2 1 x 

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= the greatest integer less than or equal to x.

Greatest Integer Function

5 5.2 51



The greatest integer function pairs the input with the greatest integer less than or equal to that

input.

–5

0 0.2

1

0

3 3.2 31

8

3

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2.2 The Algebra of Functions

 Find the sum, the difference, the product, and the quotient of two functions, and determine the domains of the resulting functions.

 Find the difference quotient for a function.

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Sums, Differences, Products, and Quotients of Functions

If f and g are functions and x is in the domain of each function, then

( )( ) ( ) ( )

( / )( ) ( ) / ( ), provided ( ) 0

  

  

 

 

f g x f x g x f g x f x g x fg x f x g x

f g x f x g x g x

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Slide 2.1-14

Example

Given that f(x) = x + 2 and g(x) = 2x + 5, find each of the following.

a) (f + g)(x) b) (f + g)(5)

Solution:

a)

b) We can find (f + g)(5) provided 5 is in the domain of each function. This is true.

f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15 (f + g)(5) = f(5) + g(5) = 7 + 15 = 22 or (f + g)(5) = 3(5) + 7 = 22

( )( ) ( ) ( )

2 2 5

3 7

f g x f x g x

x x

x

  

   

 

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Another Example

Given that f(x) = x² + 2 and g(x) = x  3, find each of the following.

a) The domain of f + g, f  g, fg, and f/g b) (f  g)(x)

c) (f/g)(x) Solution:

a) The domain of f is the set of all real numbers. The domain of g is also the set of all real numbers. The

domains of f + g, f  g, and fg are the set of numbers in the intersection of the domains—that is, the set of

numbers in both domains, or all real numbers.

For f/g, we must exclude 3, since g(3) = 0.

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Slide 2.1-16

Another Example continued

b) (f  g)(x) = f(x)  g(x) = (x² + 2)  (x  3)

= x²  x + 5 c) (f/g)(x) =

Remember to add the stipulation that x  3, since 3 is not in the domain of (f/g)(x).

2

( / )( ) ( )

( ) 2 3 f g x f x

g x x

x



 



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Difference Quotient

( ) ( )

f x h f x h

 

The ratio below is called the difference quotient, or average rate of change.

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Slide 2.1-18

Example

For the function f given by f (x) = 5x  1, find the difference quotient

Solution: We first find f(x + h):

( )

5( ) 1

5 5 1

f x h x h

x h



  

( ) ( )

f x h f x . h

 

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Example continued

( ) ( )

5 5 1 (5 1)

5 5

f x h f x h

x h x

h h

h

 

   



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Slide 2.1-20

Another Example

For the function f given by f (x) = x² + 2x  3, find the difference quotient.

Solution: We first find f(x + h):

2

2 2

( )

( ) 2( ) 3

2 2 2 3

f x h

x h x h

x xh h x h



    

     

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Example continued

2 2 2

2

( ) ( )

2 2 2 3 ( 2 3)

2 2 2 3 2 3

2 2

(2 2)

2 2

f x h f x h

x xh h x h x x

h

x xh h x h x x

h

xh h h

h h x h

x h h

 

       



       



 



     

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2.3 The Composition of Functions

 Find the composition of two functions and the domain of the composition.

 Decompose a function as a composition of two functions.

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Composition of Functions

Definition:

The , the of

and , is defined as

( )( ) ( ( )), where is in the d

composit

omain of e

and ( ) is in the domai

function composition

n of .



f g

f g x f g x

x g g x

f

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Slide 2.1-24

Example

Given that f(x) = 3x  1 and g(x) = x² + x  3, find:

a) b)

a)

( f g x)( )

2

2 2

2

( )( ) ( ( )) ( )

3( ) 1

3

3 3 9 1

3 10

3

x x

x

f g x f g x f

x x

x

 

 

  







 



(g f )( )x

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Example

a) b)

b)

( f g x)( )

2 2

2

( )( ) ( ( )) ( )

( ) (

3 1 3 1 3 1) 3

9 6 1 3 1 3

9 3 3

g f x g f x g

x x

x x x

x x

x

 

  

    





 



(g f )( )x

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Slide 2.1-26

Example

a) b)

a)

( f g)(2)

2

( )(2) ( (2))

( )

3( ) 1 8 2 2 3 3

3

f g f g

f f









(g f )(2)

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Example

a) b)

b)

( f g)(2)

2

3(2) ( )(2) ( (2))

( )

( ) ( ) 3 7

1

5 2

5

g f g f

g 



   

(g f )(2)

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Slide 2.1-28

Example

Given , find the domain of

Solution:

f(x) is not defined for negative radicands. Since the inputs of are the outputs of g, the

domain of consists of all the values in the domain of g for which g(x) is nonnegative.

( f g x)( ).

( ) 0

2 3 0

3 / 2 g x

x x



 

 

( ) 4 and ( ) 2 3 f x  x g x  x 

f g f g

The domain is { |x x  3/ 2}, or [ 3/ 2, ). 

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Decomposing a Function as a Composition

In calculus, one needs to recognize how a function can be expressed as the composition of two

functions. This can be thought of as

―decomposing‖ the function.

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Slide 2.1-30

Example

If h(x) = (3x  1)⁴, find f(x) and g(x) such that Solution: The function h(x) raises (3x  1) to the

fourth power. Two functions that can be used for the composition are:

f(x) = x⁴ and g(x) = 3x  1.

( ) ( )( ).

h x  f g x

 

⁴

( ) ( )( ) ( ( )) (3 1)

3 1

h x f g x f g x

f x x

 

 

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2.4 Symmetry and Transformations

 Determine whether a graph is symmetric with respect to the x-axis, the y-axis, and the origin.

 Determine whether a function is even, odd, or neither even nor odd.

 Given the graph of a function, graph its

transformation under translations, reflections, stretchings, and shrinkings.

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Slide 2.1-32

Symmetry

Algebraic Tests of Symmetry

x-axis: If replacing y with y produces an equivalent equation, then the graph is symmetric with respect to the x-axis.

y-axis: If replacing x with x produces an equivalent equation, then the graph is symmetric with respect to the y-axis.

Origin: If replacing x with x and y with y produces an equivalent equation, then the graph is

symmetric with respect to the origin.

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Example

Test x = y² + 2 for symmetry with respect to the x-axis, the y-axis, and the origin.

x-axis: We replace y with y:

The resulting equation is equivalent to the

original so the graph is symmetric with respect to the x-axis.

2 2

( ) 2

2

x y

 



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Slide 2.1-34

Example continued

Test x = y² + 2 for symmetry with respect to the x-axis, the y-axis, and the origin.

y-axis: We replace x with x:

The resulting equation is not equivalent to the original equation, so the graph is not symmetric with respect to the y-axis.

2 2

2 2 x y

x y



 

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Example continued

Origin: We replace x with

x and y with y:

The resulting equation is not equivalent to the

original equation, so the graph is not symmetric with respect to the origin.

2

2 2

2

( ) 2

2

x y

 

 







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Slide 2.1-36

Even and Odd Functions

If the graph of a function f is symmetric with respect to the y-axis, we say that it is an even function. That is, for each x in the domain of f, f(x) = f(x).

If the graph of a function f is symmetric with respect to the origin, we say that it is an odd function. That is, for each x in the domain of f, f(x) = f(x).

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Example

Determine whether the function is even, odd, or neither.

1.

We see that h(x) = h(x).

Thus, h is even.

4 2

( ) 4

h x  x  x

4 2

( ) ( ) 4( )

4 h x

x x





 

y = x⁴  4x²

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Slide 2.1-38

Example

Determine whether the function is even, odd, or neither.

2.

4 2

( ) 4

h x  x  x

4 2

( ) ( 4 )

4

h x x x

x x

   

  

We see that h(x)  h(x).

Thus, h is not odd.

y = x⁴  4x²

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Vertical Translation

Vertical Translation For b > 0,

the graph of y = f(x) + b is the graph of y = f(x)

shifted up b units;

the graph of y = f(x)  b is the graph of y = f(x)

shifted down b units.

y = 3x²– 3 y = 3x² y = 3x²+2

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Slide 2.1-40

Horizontal Translation

Horizontal Translation For d > 0,

the graph of y = f(x  d) is the graph of y = f(x)

shifted right d units;

the graph of y = f(x + d) is the graph of y = f(x)

shifted left d units.

y = 3x²

y = (3x – 2)²

y = (3x + 2)²

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Reflections

The graph of y = f(x) is the reflection of the graph of y = f(x) across the x-axis.

The graph of y = f(x) is the reflection of the graph of y = f(x) across the y-axis.

If a point (x, y) is on the graph of y = f(x), then (x, y) is on the graph of y = f(x), and (x, y) is on the graph of y = f(x).

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Slide 2.1-42

Example

Reflection of the graph y = 3x³  4x² across the x-axis.

y = 3x³  4x²

y = 3x³ + 4x²

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Example

Reflection of the graph y = x³  2x² across the y-axis.

y = x³  2x² y = -x³ + 2x²

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Slide 2.1-44

Vertical Stretching and Shrinking

The graph of y = af(x) can be obtained from the graph of y = f(x) by

stretching vertically for |a| > 1, or shrinking vertically for 0 < |a| < 1.

For a < 0, the graph is also reflected across the x- axis.

(The y-coordinates of the graph of y = af(x) can be obtained by multiplying the y-coordinates of y = f(x) by a.)

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Examples

Stretch y = x³ – x vertically.

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Slide 2.1-46

Examples

Shrink y = x³ – x vertically.

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Examples

Stretch and reflect y = x³ – x across the x -axis

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Slide 2.1-48

Horizontal Stretching or Shrinking

The graph of y = f(cx) can be obtained from the graph of y = f(x) by

shrinking horizontally for |c| > 1, or stretching horizontally for 0 < |c| < 1.

For c < 0, the graph is also reflected across the y- axis.

(The x-coordinates of the graph of y = f(cx) can be obtained by dividing the x-coordinates of the graph of y = f(x) by c.)

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Examples

Shrink y = x³ – x horizontally.

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Slide 2.1-50

Examples

Stretch y = x³ – x horizontally.

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Examples

Stretch horizontally and reflect y = x³ – x.

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2.5 Variation and Applications

 Find equations of direct, inverse, and combined variation given values of the variables.

 Solve applied problems involving variation.

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Direct Variation

If a situation gives rise to a linear function f(x) = kx, or y = kx, where k is a positive

constant, we say that we have direct variation, or that y varies directly as x, or that y is

directly proportional to x. The number k is called the variation constant, or constant of proportionality.

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Slide 2.1-54

Direct Variation

The graph of y = kx, k > 0, always goes through the origin and rises from left to right. As x increases, y increases; that is, the function is increasing on the interval (0,). The constant k is also the slope of the line.

, 0 y  kx k 

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Direct Variation

Example: Find the variation constant and an equation of

variation in which y varies directly as x, and y = 42 when x = 3.

Solution: We know that (3, 42) is a solution of y = kx.

y = kx 42 = k  3

14 = k

The variation constant 14, is the rate of change of y with respect to x. The equation of variation is y = 14x.

42 3  k

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Slide 2.1-56

Application

Example: Wages. A cashier earns an hourly wage. If the cashier worked 18 hours and earned $168.30, how much will the cashier earn if she works 33 hours?

Solution: We can express the amount of money earned as a function of the amount of hours worked.

I(h) = kh I(18) = k  18

$168.30 = k  18

$9.35 = k The hourly wage is the variation constant.

Next, we use the equation to find how much the cashier will earn if she works 33 hours.

I(33) = $9.35(33)

= $308.55

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Inverse Variation

If a situation gives rise to a function f(x) = k/x, or y = k/x, where k is a positive constant, we say that we

have inverse variation, or that y varies inversely as x, or that y is inversely proportional to x. The

number k is called the variation constant, or constant of proportionality.

For the graph y = k/x, k  0, as x increases, y

decreases; that is, the function is decreasing on the interval (0, ).

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Slide 2.1-58

Inverse Variation

For the graph y = k/x, k  0, as x increases, y

decreases; that is, the function is decreasing on the interval (0, ).

, 0 y k k

 x 

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Inverse Variation

Example: Find the variation constant and an equation of variation in which y varies inversely as x, and y = 22

when x = 0.4.

Solution:

The variation constant is 8.8. The equation of variation is y = 8.8/x.

22 0.4 (0.4)22

8.8 y k

x k k k



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Slide 2.1-60

Application

Example: Road Construction. The time t required to do a job varies inversely as the number of people P who work on the job (assuming that they all work at the same rate). If it takes 180 days for 12 workers to complete a job, how long will it take 15 workers to complete the same job?

Solution: We can express the amount of time required, in days, as a function of the number of people working.

t varies inversely as P

This is the variation constant.

( )

(12) 12 180 12 2160

t P k

P t k

k k



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Application continued

The equation of variation is t(P) = 2160/P.

Next we compute t(15).

It would take 144 days for 15 people to complete the same job.

( ) 2160 (15) 2160

15 144

t P P

t



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Slide 2.1-62

Combined Variation

Other kinds of variation:

y varies directly as the nth power of x if there is some positive constant k such that .

y varies inversely as the nth power of x if there is some positive constant k such that .

y varies jointly as x and z if there is some positive constant k such that y = kxz.

y  kxn

n

y k

 x

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Example

The luminance of a light (E) varies directly with the intensity (I) of the light and inversely with the square distance (D) from the light. At a distance of 10 feet, a light meter reads 3 units for a 50-cd lamp. Find the luminance of a 27-cd lamp at a distance of 9 feet.

Solve for k.

Substitute the second set of data into the equation.

2

3 50

10 6

E k I D k k



 



2

6 27 9 2 E E

 



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2.6 Solving Linear Inequalities

 Solve linear inequalities.

 Solve compound inequalities.

 Solve inequalities with absolute value.

 Solve applied problems using inequalities.

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Inequalities

An inequality is a sentence with <, >, , or  as its verb.

Examples: 5x  7 < 3 + 4x 3(x + 6)  4(x  3)

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Slide 2.1-66

Principles for Solving Inequalities

For any real numbers a, b, and c:

The Addition Principle for Inequalities: If a < b is true, then a + c < b + c is true.

The Multiplication Principle for Inequalities: If a < b and c > 0 are true, then ac < bc is true. If a < b and c < 0, then ac > bc is true.

Similar statements hold for a  b.

When both sides of an inequality are multiplied or divided by a negative number, we must reverse the inequality sign.

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Examples

Solve:

{x|x < 2} or (, 2)

Solve:

{x|x  4} or [4, )

4 6 2 10

4 2 4

2 4

2

x x

  

 

6( 3) 7( 2)

6 18 7 14

4 4

x x

  

 

 

–5 0 5

)

–5 0 5

[

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Slide 2.1-68

Compound Inequalities

When two inequalities are joined by the word and or the word or, a compound inequality is formed.

Conjunction contains the word and.

Example: 7 < 3x + 5 and 3x + 9  6 Disjunction contains the word or.

Example: 3x + 5  6 or 3x + 6 > 12

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Examples

Solve: Solve:

4x  5  3 or 4x  5 > 3

4 3x 8 11

   

4 3 8 11

12 3 3

4 1

x x x

   

  

4 5 3 4 5 3

4 2 4 8

1 2

2

x or x

x x

    

 

0

–5 5

( ]

0

–5 5

] (

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Slide 2.1-70

Inequalities with Absolute Value

Inequalities sometimes contain absolute-value

notation. The following properties are used to solve them.

For a > 0 and an algebraic expression X:

|X| < a is equivalent to a < X < a.

|X| > a is equivalent to X < a or X > a.

Similar statements hold for |X|  a and |X|  a.

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Example

Solve:

4 1 3

3 4 1 3

4 4 2

1 1

2 x

x x x

 

   

  

0

–5 5

( )

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Slide 2.1-72

Application

Johnson Catering charges $100 plus $30 per hour to cater an event. Catherine’s Catering charges

$50 per hour. For what lengths of time does it cost less to hire Catherine’s Catering?

1. Familiarize. Read the problem.

2. Translate. Catherine’s is less than Johnson 50x < 100 + 30x

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Application continued

3. Carry out.

4. Check.

5. State. For values of x < 5 hr, Catherine’s Catering

50 100 30 20 100

5

x x

 



50(5) ? 100 30(5) 250 ? 100 150 250 250

 

