Answers & Hints for Selected Exercises 669- Revise the proof of Part (a), making appropriate changes.
- (a) Increasing on (-2, ~) and (3, +oo); decreasing on (-oo, -2) and(~, 3).
Local minimum y = 0 at x = -2, 3; local maximum y =^2 ; at x = ~-
(c) Increasing on (-oo,-1) and (-1,0); decreasing on (0,1) and (1,oo). No
local minimum; local maximum y = -1 at x = 0.
(e) Increasing on (-oo, -0); decreasing on (0, oo). No local minimum; local
maximum y = 1 at x = 0.
(f) Increasing on (-1,1); decreasing on (-oo,-1) and (1,oo). Local mini-
mum y = -~ at x = -1; local maximum y = ~ at x = 1. - (a) f is differentiable everywhere, since
x =/:- 0 =? f^1 ( x) = 1 + 4x sin ~ - 2 cos ~, and
+2 2 · I
f' (0) = lim x x sm-: = lim (1 + 2x sin l) = 1.
h~o x h~o x
(b) Let I be any neighborhood of 0. Then I contains a tail of the sequence
{ n^1 .,,.} since n^1 .,,. --+ 0. Note that f' ( n^1 .,,.) = 1 + n4... sin mr - 2 cos mr = -1 if n
is even, 3 if n is odd. Thus, I contains points x where f'(x) < 0 and points x
where f'(x) > 0. Apply Thm. 6.3.5. - Revise the proof of Case 1, changing inequalities appropriately.
- Suppose f is differentiable on an open interval I and f'(x) =/:- 0 on I.
Suppose 3a, b EI 3 f'(a) < 0 and f'(b) > 0. Then, by Thm. 6.3.7, :Jc between
a and b such that f'(c) = 0. Contradiction.
EXERCISE SET 6.4- Given: d E [a, b] and f(d) > J(a) = J(b). Then d =/:-a, b, so d E (a, b) and
f(d) = maxf[a, b]. That is, f has a local max at d. .-. by Thm. 6.3.4, f'(d) = 0. - Let x 1 =/:-X2 in I , say x1 < x2. If f (x1) = f (x2) then by Rolle's Thm. applied
to [x1,x2], :Jc E (x1,x2) 3 f'(c) = 0. But Ve E J, f'(c) =/:-0. :. f(x1) =/:-f(x2). - Let f(x) = 3x^5 - 2x^3 + 12 x - 8. Then f'(x ) = 15 x^4 - 6x^2 + 12, a quadratic
expression in x^2 with discriminant D < 0. Thus, ':::/ x E JR 3 f'(x) = 0. Hence,
by Rolle's Thm., ':::/ x 1 ,x2 E JR 3 f(x 1 ) = f(x2); that is, f is 1-1 on R - The function f(x) = 7x^3 - 5x^2 + 4x - 10 is continuous on JR, f(O) < 0, and
f(2) > 0, so by the intermediate value theorem, :Jc E (0, 2) 3 f(c) = 0. Thus,
the equation f(x ) = 0 has at least one root.
Suppose 3 x 1 < x2 3 f (xi) = 0 and f(x2) = 0. Then by Rolle's Thm.,
:Jc E (x 1 , x 2 ) 3 f'(c) = 0. But f'(x) = 21x^2 -10x+4, a quadratic with negative
discriminant. So ':::/ c E JR 3 f'(c) = 0. Contradiction. .-. The equation f(x) ;= 0
cannot have more than one root.