328 Chapter 6 • Differentiable Functions
- By Exercise 21 , if f is differentiable and has a continuous derivative in
some neighborhood of xo, and f'(xo) =f 0, then f is strictly monotone
on some neighborhood of x 0. Resolve the apparent contradiction between
this result and Example 6.3.6. - Suppose f and g are differentiable on the interval [a, +oo), j(a) ::; g(a),
and 't:/x > a, f'(x) < g'(x). Prove that 't:/x > a, j(x) < g(x). [Hint: let
h(x) = g(x) - f(x).] (This remains true if "<" is replaced by "S" in
both places.) - Prove that 't:/x > 1, ex >ex. [See also Exercise 28.]
- Prove the first derivative test: Suppose f is continuous at x 0 and
differentiable in a deleted neighborhood NJ;(x 0 ), for some 8 > 0. Then
(a) if 't:/x E (xo -5,xo), f'(x) 2'. 0 and 't:/x E (xo, xo + 8), f'(x) ::; 0, then
f has a local maximum at c.
(b) if 't:/x E (xo - 8, xo), f'(x) ::; 0 and 't:/x E (xo, xo + 5), f'(x) 2'. 0, then
f has a local minimum at c.
26. Prove the second derivative test: Suppose f is differentiable on a
neighborhood of xo, f'(xo) = 0, and f"(xo) exists and is nonzero. Then
(a) if f"(xo) < 0, then f has a lo cal maximum at xo.
(b) if f" ( xo) > 0, then f has a local minimum at x 0.- Prove the following theorem that is frequently used in elementary calcu-
lus, but rarely proved there. Suppose f is differentiable on an open interval
I , f has a local maximum (minimum) at xo EI, and x 0 is the only point
of I where j' (x) = 0. Then f has its absolute maximum (minimum) value
for I at xo. - Use Exercise 27 to prove that 't:/x =f 0, ex > ex. [See Exercise 24.]
- Use the mean value theorem to prove Theorem 6.1.14.
30.31.1 1 + lxl
Prove that the functions f(x) = - and g(x) = --have the same
x x
derivative but do not differ by a constant. Does this contradict Corollary
6.4.5? Explain.Derivatives (unlike continuous functions) are not necessarily bounded on
compact sets. Show that there is a compact interval on which the func-tion f(x) = { x
2
sin(l/x2
) ~f x =f O} is differentiable but on which the
0 if x = 0
derivative f' is unbounded.