Solução_Calculo_Stewart_6e

F.
192 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
33. f(x) =cx +sinx ⇒ f 0 (x) =c +cosx ⇒ f 00 (x) =− sin x
f(−x) =−f(x),sof is an odd function and its graph is symmetric with respect to the origin.
f(x) =0 ⇔ sin x = −cx,so0 is always an x-intercept.
f 0 (x) =0 ⇔ cos x = −c, so there is no critical number when |c| > 1. If|c| ≤ 1, then there are infinitely
many critical numbers. If x 1 is the unique solution of cos x = −c in the interval [0,π], then the critical numbers are 2nπ ± x 1 ,
where n ranges over the integers. (Special cases: When c =1, x 1 =0;whenc =0, x = π ;andwhenc = −1, x 2 1 = π.)
f 00 (x) < 0 ⇔ sin x>0, sof is CD on intervals of the form (2nπ, (2n +1)π). f is CU on intervals of the form
((2n − 1)π, 2nπ). Theinflection points of f are the points (2nπ, 2nπc),wheren is an integer.
If c ≥ 1, thenf 0 (x) ≥ 0 for all x,sof is increasing and has no extremum. If c ≤−1,thenf 0 (x) ≤ 0 for all x,sof is
decreasing and has no extremum. If |c| < 1, thenf 0 (x) > 0 ⇔ cos x>−c ⇔ x is in an interval of the form
(2nπ − x 1 , 2nπ + x 1 ) for some integer n. These are the intervals on which f is increasing. Similarly, we
find that f is decreasing on the intervals of the form (2nπ + x 1 , 2(n +1)π − x 1 ).Thus,f has local maxima at the points
2nπ + x 1,wheref has the values c(2nπ + x 1)+sinx 1 = c(2nπ + x √ 1)+ 1 − c 2 ,andf has local minima at the points
2nπ − x 1,wherewehavef(2nπ − x 1)=c(2nπ − x 1) − sin x 1 = c(2nπ − x 1) − √ 1 − c 2 .
The transitional values of c are −1 and 1. Theinflection points move vertically, but not horizontally, when c changes.
When |c| ≥ 1, there is no extremum. For |c| < 1, the maxima are spaced
2π apart horizontally, as are the minima. The horizontal spacing between
maxima and adjacent minima is regular (and equals π)whenc =0,but
the horizontal space between a local maximum and the nearest local
minimum shrinks as |c| approaches 1.
35. If c0,then
lim
x→−∞
If c =0,thenf(x) =x,so
lim
x→−∞ xe−cx =
lim
x→−∞
f(x) =−∞,and lim f(x) H 1
= lim =0.
x→∞ x→∞ cecx lim
x→±∞
f(x) =±∞, respectively.
x H 1
= lim =0,and lim f(x) =∞.
e cx x→−∞ cecx x→∞
So we see that c =0is a transitional value. We now exclude the case c =0, since we know how the function behaves
in that case. To find the maxima and minima of f, we differentiate: f(x) =xe −cx
⇒
f 0 (x) =x(−ce −cx )+e −cx =(1− cx)e −cx .Thisis0 when 1 − cx =0 ⇔ x =1/c. Ifc