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Transformation of a Piecewise Function

Module by: Debbie Trahan

Summary: In this module students will investigate transformations on a piecewise function.

In this module, you will investigate the transformations on a piecewise function.

The graph of the function f f shown in figure 1 consists of two line segments.

Figure 1
Figure 1 (pieprob.png)

Use function f f shown in figure 1 to answer the questions in this lesson.

Exercise 1

Find f-1 f -1 .

Solution 1

Read the point from the graph. f-1=0 f -1 0

Exercise 2

Estimate f1.25 f 1.25 from the graph then find the exact answer algebraically.

Solution 2

Estimated answer will vary.

Write an equation for f f when 0<x<2 0 x 2

y=-3x+3 y -3 x 3

Substitute 1.25 in place of x x and solve for y y .

y=-3×1.25+3 y -3 1.25 3

f1.25=-0.75 f 1.25 -0.75

Exercise 3

Write a piecewise function for f f .

Solution 3

fx=3x+3if-2≤x<0-3x+3if0≤x≤2 f x 3 x 3 -2 x 0 -3 x 3 0 x 2

Exercise 4

Find the value of x x when fx=-2 f x -2 .

Solution 4

x=-53 x -5 3 or x=53 x 5 3

Exercise 5

For what values of x x is f f increasing?

Solution 5

-20 -2 0

Exercise 6

Find the domain of f f .

Solution 6

-22 -2 2

Exercise 7

Find the minimum value of f f .

Solution 7

-3 -3

Exercise 8

Find the maximum value of f f .

Solution 8

3 3

Exercise 9

Find the area of the region beneath f f in the first quadrant.

Solution 9

1.5 1.5 square units

Exercise 10

Find the area of the region bounded by f f , x=0 x 0 , x=12 x 1 2 , and y=0 y 0 .

Solution 10

1.125 1.125 square units

Exercise 11

Part a: Graph fx-1 f x 1 then find the maximum and the minimum value of fx-1 f x 1 .

Part b: Find the area of the region beneath fx-1 f x 1 in the first quadrant.

Solution 11

Part a:

Figure 2
Figure 2 (pansk2.png)

The maximum value is 3 3 .

The miminimum value is -3 -3 .

Part b: 3 3 square units .

Exercise 12

Part a: Graph fx+2 f x 2 then find the x - intercepts of the graph and the maximum and the minimum value of fx+2 f x 2 .

Part b: Find the area of the region beneath fx+2 f x 2 in the first quadrant.

Solution 12

Part a:

Figure 3
Figure 3 (pansl.png)

The x - intercepts of the graph are -53 -5 3 and 53 5 3 .

The maximum value is 5 5

The minimum value is -1 -1

Part b: 256 25 6 square units .

Exercise 13

Graph -fx f x .

Solution 13

Figure 4
Figure 4 (m1.gif)

Exercise 14

Graph 13fx 1 3 f x .

Solution 14

Figure 5
Figure 5 (m2.gif)

Exercise 15

Graph f2x f 2 x .

Solution 15

Figure 6
Figure 6 (m3.gif)

Exercise 16

Graph f2x+2 f 2 x 2 .

Solution 16

Figure 7
Figure 7 (m4.gif)

Exercise 17

Graph -fx+1 f x 1 .

Solution 17

Figure 8
Figure 8 (m5.gif)

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