670 Appendix C • Answers & Hints for Selected Exercises- Suppose f"(x) = 0 for all x ER By Thm. 6.4.4, f'(x) must be constant;
say f'(x) = c for all x ER Let g(x) =ex. Since 't/x E JR, f'(x) = g'(x), Cor.
6.4.5 says that :3d E JR 3 't/x E JR, f(x) = g(x) + d. That is, f(x) =ex+ d. - Suppose that 't/x, y E JR, If (x)-f (y)I ::; Ix -yl^2. Then I f(x;:::;(y) I ::; Ix -yl,
so Vxo E JR, lim I f(x):::J(xo) I ::; lim Ix - xol = 0. That is , Vxo E JR, f'(x) = 0.
X---+Xo x xo X---+Xo
. ·. By Thm. 6.4.4, f is constant on R
13. Revise the proof of (a), making appropriate changes.
15. Let f(x) = x^3 and g(x) = -x^3 on I= (-1, 1). Then f is strictly increasing
on I, g is strictly decreasing on I, and 0 EI, yet f'(O) :f 0 and g'(O) f:. 0.
17. Let x E (0 , ~).Then tanx is differentiable on [O,x] so by the MVT, ::Jc E
(0, x) 3 tan' ( c) = tan ~=~an °; i.e., sec^2 c = ta~ x. Since J sec cj > 1, this says that
't/x E (0, ~), ta~x > l ; i.e., tanx > x.
- Let x > 1 and f(x ) = lnx. Applying the MVT to f on [l,x], ::Jc E (1,x) 3
l c = ln' c = Inx-1 x-Inl =~-Since x-1 1 < c < x ' l x < l c < (^1) ' sol<~< x x-1 1. Since
x - 1 > 0, this implies x~l < lnx < x - 1.
- Suppose f' is continuous at some interior point x 0 of its domain, and
f'(x 0 ) > 0. By the "nbd inequality property of continuous functions,'' (Exercise
5.1.26) :38 > 0 3 't/x E N 0 (xo), f'(x) > 0. Then by Thm. 6.4.6 (c), f is strictly
increasing on N 0 (xo). Similarly for f'(xo) < 0. - Define h : [a, oo) --> JR by h(x) = g(x) - f(x). Then 't/x > a, h'(x) =
g'(x) - f'(x) > 0, so by Thm. 6.4.6, his strictly increasing on [a, oo). Then,
't/x >a, h(x ) > h(a); i.e., g(x) - f(x) > g(a) - f(a) ~ O; :. g(x) > f(x). - (a) Suppose f'(x) ~ 0 in (xo-8,xo) and f'(x)::; 0 in (x 0 ,x 0 +5). (i) Let
x E (xo - 8,xo). By the MVT applied to f on [x,xo], ::Jc E (x,xo) 3 f'(c) =
f(x:~:::;(x). Since f'(c) ~ 0, and x 0 - x > 0, we must have f(xo) - f(x) ~ O;
i.e., f(xo) ~ f(x). (ii) Similarly, x E (xo, xo + 8) =? by MVT, :3d E (x 0 , x) 3
f'(d) = f(x;:::;~xo) :S: 0 =? f(x) :S: f(xo).
Therefore, f (xo) =max f (xo -8, xo + 8), so f has a local maximum at x 0. - Assume the hypotheses. Then :38 > 0 3 N 0 ( Xo) <;;; I and f ( xo) = max f ( N 0 ( xo)).
Suppose f is not the absolute max off on I. Then :3x 1 EI 3 f(x1) > f(xo).
Case 1: X1 > xo. By the MVT, ::Jc E (x 0 , x 1 ) 3 f'(c) = f(xi)-f(xi-xo xo) > - Since f'(x)-=!= 0 in I, f is not constant on (xo,min{xo+8,c}), so:3x2 E
(xo, min{xo + 8, c}) 3 f(x2) -=!= f(xo). Then f(x2) < f(xo) since f(xo) =
max f (xo-8, xo+8). Then by the MVT, :3d E (x 0 , x 2 ) 3 f'(d) = f(xx2-xo^2 )-f(xo) < O.
But then f'(d) < 0 < f'(c), so by the intermediate value property of
derivatives (Thm. 6.3.7) :3e between c and d 3 f'(e) = 0. Recall that c, d > 0,
so e is a second point of I at which f' = 0. Contradiction.:. f(x 0 ) = maxf(I).