College Algebra 9th txtbk

246 CHAPTER 3 FUNCTIONS
Step 2. Replace f(x) with y, then interchange x and y.
x 4y y 2
Step 3. Solve the equation for y.
x 4y y 2
y 2 4y x
y 2 4x 4 x 4
(y 2) 2 4 x
y 2 14 x
y 2 14 x
y 4x x 2
Rewrite so that the coefficient of y 2 is 1.
Add 4 to both sides to complete the square.
Factor the left side.
Take the square root of both sides.
Add 2 to both sides.
Now we have two possible solutions. The domain of f was (– , 2], and this should
be the range of f 1 . In other words, the output of the inverse is never greater
than 2. But y 2 14 x would always be greater than or equal to 2, so we
must instead choose y 2 14 x.
5
y
y f(x)
y x
f 1 (x) 2 14 x
Step 4. The domain of f 1 is the range of f. We can see from Figure 10 that this is
(, 4]. Notice that the equation we found for f 1 (x) is defined for these values.
Our final answer is
y f 1 (x)
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x
f 1 (x) 2 14 x
The check is again left for the reader.
x 4
Z Figure 11
The graphs of f, f 1 , and y x are shown in Figure 11. To aid in graphing f 1 ,
we plotted several points on the graph of f and then reflected these points in the line
y x.
MATCHED PROBLEM 5
Find the inverse of f(x) 4x x 2 , x 2. Graph f, f 1 , and y x in the same coordinate
system.
Technology Connections
To reproduce Figure 11 on a graphing calculator, first enter
y 1 (4x x 2 )(x 2)
in the equation editor (Fig. 12) and graph (Fig. 13). (For
graphs involving both f and f 1 it is best to use a squared
viewing window.) The Boolean expression (x 2) is
assigned the value 1 if the inequality is true and 0 if it is
false. The calculator recognizes that division by 0 is an undefined
operation and no graph is drawn for x 2. Now enter
y 2 2 14 x
and
in the equation editor and graph (Fig. 14).
y 3 x
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Z Figure 12 Z Figure 13 Z Figure 14
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