Do These Quickly
Q1. In the composite function m (d (x)), function d is called the function.
Q2. In the composite function m (d (x)), function m is called the function.
Q3. Give another symbol for m (d (x)).
DEFINITIONS AND PROPERTIES: Function Inverses
• The inverse of a relation in two variables is formed by interchanging the two variables.
• If the inverse of function f is also a function, then f is invertible.
• If f is invertible and y = f(x), then you can write the inverse of f as y = f –1(x).
• To plot the graph of the inverse of a function, either
Interchange the variables, solve for y, and plot the resulting equation(s), or
Use parametric mode, as in Example 2.
• If f is invertible, then the compositions of f and f –1 are
f –1(f (x)) = x, provided x is in the domain of f and f (x) is in the domain of f–1
f(f –1(x)) = x, provided x is in the domain of f –1and f –1(x) is in the domain of f
• A one-to-one function is invertible. Strictly increasing or strictly decreasing functions are one-to-one functions.
Q4. If f (x) = 2x and g(x) = x + 3, find f (g (1)). horizontal translation of f (x) by 3 units.
For Problems 1–4, sketch the line y = x and sketch the inverse relation on a copy of the given figure. Be sure that the inverse relation is a reflection of the function graph across y = x. Tell whether or not the inverse relation is a function.
1.
2.
3.
|
3 –5|
=4.
For Problems 5–16, plot the function in the given domain using parametric mode. On the same screen, plot its inverse relation. Tell whether or not the inverse relation is a function. Sketch the graphs.
5. f (x) = 2x –6, –1 x 5 For Problems 17–20, write an equation for the inverse relation by interchanging the variables and solving for y in terms of x. Then plot the function and its inverse on the same screen, using function mode. Sketch the result, showing that the function and its inverse are reflections across the line y = x.
Tell whether or not the inverse relation is a function.
23. Cost of Owning a Car Problem: Suppose you have fixed costs (car payments, insurance, and so on) of $300 per month and operating costs of $0.25 per mile you drive. The monthly cost of owning the car is given by the linear function
c(x) = 0.25x + 300
where x is the number of miles you drive the car in a given month and c(x) is the number of dollars per month you spend.
a. Find c(1000). Explain the real-world meaning of the answer.
b. Find an equation for c–1(x), where x now stands for the number of dollars you spend instead of the number of miles you drive.
Explain why you can use the symbol c–1 for the inverse relation. Use the equation of c–1(x) to find c–1(437) and explain its real-world meaning.
c. Plot y1 = c(x) and y2 = c–1(x) on the same screen, using function mode. Use a window of 0 x 1000 and use equal scales on the two axes. Sketch the two graphs, showing how they are related to the line y = x.
24. Deer Problem: The surface area of a deer's body is approximately proportional to the
power of the deer’s weight. (This is true because the area is proportional to the square of the length and the weight is proportional to the cube of the length.) Suppose that the particular equation for area as a function of weight is given by the power function
A(x) = 0.4x2/3
where x is the weight in pounds and A(x) is the surface area measured in square feet.
a. Find A(50), A(100), and A(150). Explain the real-world meaning of the answers.
b. True or false: “A deer twice the weight of another deer has a surface area twice that of the other deer.” Give numerical evidence to support your answer.
25. Braking Distance Problem: The length of skid marks, d(x) feet, left by a car braking to a stop is a direct square power function of x, the speed in miles per hour when the brakes were applied. Based on information in the Texas Drivers Handbook (1997), d(x) is given approximately by
d(x) = 0.057x2 for x 0
The graph of this function is shown in Figure 1-5h.
Figure 1-5h
a. When police officers investigate automobile represent in the context of this problem?
c. Suppose that the domain of function d started at –20 instead of zero. With your grapher in parametric mode, plot the graphs of function d and its inverse relation. Use the window shown in Figure 1-5h with a t-range of –20 t 70. Sketch the result.
d. Explain why the inverse of d in part c is not a function. What relationship do you notice between the domain and range of d and its inverse?
the function is increasing in some parts of the domain and decreasing in other parts.
27. Tabular Function Problem: A function is defined by the following table of values. Tell whether or not the function is invertible.
50 1000
60 2000
70 2500
80 2000
90 1500
28. Horizontal Line Test Problem: The vertical line test of Problem 39 in Section 1-2 helps you see graphically that a relation is a function if no vertical line crosses the graph more than once.
The horizontal line test allows you to tell whether or not a function is invertible. Sketch two graphs, one for an invertible function and one for a noninvertible function, that illustrate the following test.
Figure 1-5i
In Section 1-3 you learned that if y = f(x), then multiplying x by a constant causes a horizontal dilation. Suppose that the constant is –1. Each x-value will be 1/(–1) or –1 times what it was in the pre-image. Figure 1-6a shows that the resulting image is a horizontal reflection of the graph across the y-axis. The new graph is the same size and shape, just a mirror image of the original.
Similarly, a vertical dilation by a factor of –1 reflects the graph vertically across the x-axis.
In this section you will learn special transformations of functions that reflect the graph in various ways. You will also learn what happens when you take the absolute value of a function or of the independent variable x. Finally, you will learn about odd and even functions.