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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.

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

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

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

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

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

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

Exercise 5

For what values of x x is f f increasing?

Solution

-20 -2 0

Exercise 6

Find the domain of f f .

Solution

-22 -2 2

Exercise 7

Find the minimum value of f f .

Solution

-3 -3

Exercise 8

Find the maximum value of f f .

Solution

3 3

Exercise 9

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

Solution

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

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

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

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

Figure 4
Figure 4 (m1.gif)

Exercise 14

Graph 13fx 1 3 f x .

Solution

Figure 5
Figure 5 (m2.gif)

Exercise 15

Graph f2x f 2 x .

Solution

Figure 6
Figure 6 (m3.gif)

Exercise 16

Graph f2x+2 f 2 x 2 .

Solution

Figure 7
Figure 7 (m4.gif)

Exercise 17

Graph -fx+1 f x 1 .

Solution

Figure 8
Figure 8 (m5.gif)

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