Solução_Calculo_Stewart_6e

F.
TX.10
28 ¤ CHAPTER 1 FUNCTIONS AND MODELS
21. 2 ft =24in, f(24) = 24 2 in = 576 in =48ft. g(24) = 2 24 in =2 24 /(12 · 5280) mi ≈ 265 mi
23. The graph of g finally surpasses that of f at x ≈ 35.8.
25. (a) Fifteen hours represents 5 doubling periods (one doubling period is three hours). 100 · 2 5 = 3200
(b) In t hours, there will be t/3 doubling periods. The initial population is 100,
so the population y at time t is y = 100 · 2 t/3 .
(c) t =20 ⇒ y = 100 · 2 20/3 ≈ 10,159
(d) We graph y 1 = 100 · 2 x/3 and y 2 =50,000. The two curves intersect at
x ≈ 26.9, so the population reaches 50,000 in about 26.9 hours.
27. An exponential model is y = ab t ,wherea =3.154832569 × 10 −12
and b =1.017764706. This model gives y(1993) ≈ 5498 million and
y(2010) ≈ 7417 million.
29. From the graph, it appears that f is an odd function (f is undefined for x =0).
To prove this, we must show that f(−x) =−f(x).
f(−x) = 1 − e1/(−x) 1 −
1 − 1
e(−1/x)
=
1+e1/(−x) 1+e = e 1/x
(−1/x)
1+ 1
e 1/x
= − 1 − e1/x
= −f(x)
1+e1/x so f is an odd function.
· e1/x
e = e1/x − 1
1/x e 1/x +1
1.6 Inverse Functions and Logarithms
1. (a) See Definition 1.
(b) It must pass the Horizontal Line Test.
3. f is not one-to-one because 2 6= 6,butf(2) = 2.0 =f(6).
5. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is one-to-one.