Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS ¤ 29
7. The horizontal line y =0(the x-axis) intersects the graph of f in more than one point. Thus, by the Horizontal Line Test,
f is not one-to-one.
9. The graph of f(x) =x 2 − 2x is a parabola with axis of symmetry x = − b
2a = − −2 =1.Pickanyx-values equidistant
2(1)
from 1 to find two equal function values. For example, f(0) = 0 and f(2) = 0,sof is not one-to-one.
11. g(x) =1/x. x 1 6=x 2 ⇒ 1/x 1 6=1/x 2 ⇒ g (x 1 ) 6=g (x 2 ),sog is one-to-one.
Geometric solution: The graph of g is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test,
so g is one-to-one.
13. A football will attain every height h up to its maximum height twice: once on the way up, and again on the way down. Thus,
even if t 1 does not equal t 2 , f(t 1 ) may equal f(t 2 ),sof is not 1-1.
15. Since f(2) = 9 and f is 1-1, we know that f −1 (9) = 2. Remember, if the point (2, 9) is on the graph of f, then the point
(9, 2) is on the graph of f −1 .
17. First, we must determine x such that g(x) =4. By inspection, we see that if x =0,theng(x) =4.Sinceg is 1-1 (g is an
increasing function), it has an inverse, and g −1 (4) = 0.
19. We solve C = 5 (F − 32) for F : 9 C = F − 32 ⇒ F = 9 9 5 5
C +32. This gives us a formula for the inverse function, that
is, the Fahrenheit temperature F as a function of the Celsius temperature C. F ≥−459.67 ⇒ 9 5 C +32≥−459.67 ⇒
9
5
C ≥−491.67 ⇒ C ≥−273.15, the domain of the inverse function.
21. f(x) = √ 10 − 3x ⇒ y = √ 10 − 3x (y ≥ 0) ⇒ y 2 =10− 3x ⇒ 3x =10− y 2 ⇒ x = − 1 3 y2 + 10
3 .
Interchange x and y: y = − 1 3 x2 + 10
3 .Sof −1 (x) =− 1 3 x2 + 10
3 . Note that the domain of f −1 is x ≥ 0.
23. y = f(x) =e x3 ⇒ ln y = x 3 ⇒ x = 3√ ln y. Interchange x and y: y = 3√ ln x. Sof −1 (x) = 3√ ln x.
25. y = f(x) =ln(x +3) ⇒ x +3=e y ⇒ x = e y − 3. Interchange x and y: y = e x − 3. Sof −1 (x) =e x − 3.
27. y = f(x) =x 4 +1 ⇒ y − 1=x 4 ⇒ x = 4√ y − 1 (not ± since
x ≥ 0). Interchange x and y: y = 4√ x − 1. Sof −1 (x) = 4√ x − 1. The
graph of y = 4√ x − 1 is just the graph of y = 4√ x shifted right one unit.
From the graph, we see that f and f −1 are reflections about the line y = x.
29. Reflect the graph of f about the line y = x. The points (−1, −2), (1, −1),
(2, 2),and(3, 3) on f are reflected to (−2, −1), (−1, 1), (2, 2),and(3, 3)
on f −1 .