Analysis Calc Proofs?

Question

I need some help, thanks!

15. a. Let f be a function such at | f(x) | =< x^2 for all x. Prove that f is differentiable at 0.
b. This result can be generalized if x^2 is replaced by | g(x) |, where g has what property?

19.a. Suppose that f(a)=g(a)=h(a), that f(x)= b. Show that the conclusion does not follow if we omit the hypothesis f(a)=g(a)=h(a).

22. Suppose that f is differentiable at x. Prove that:
f '(x) = lim (h ~> 0) [f(x+h) - f(x-h)] / 2h

28.a. Find f '(x) if f(x) = |x|^3. Find f ''(x). Does f '''(x) exist for all x?
b. Analyze f similarly if f(x) = x^4 for x >= 0 and f(x) = -x^4 for x =< 0.

More to come, anything on these would be immensely helpful. Thanks again!

~Andrew

Answer 1

15a. f'(0) = lim x->0 (f(x)-f(0))/(x-0)
= lim x->0 f(x)/x.
| f(x) / x | <= |x^2| / |x| <= |x|, so apply the squeeze theorem with bounding functions x and -x to show that f'(0) = 0

b. the key step in a was that |x^2|/|x| -> 0 as x->0

19a.
g'(a) = lim(e->0) [g(a+e)-g(a)]/e, but by the assumptions,
[f(a+e)-f(a)]/e <= [g(a+e)-g(a)]/e <= [h(a+e)-h(a)]/h, and since the upper and lower bounds converge to f'(a)=h'(a), by the squeeze theorem, g'(a) must exist and equal f'(a) and h'(a).
b. If the function values are not equal, then we cannot use f(x) <= g(x) <= h(x) to conclude that
[f(a+e)-f(a)]/e <= [g(a+e)-g(a)]/e <= [h(a+e)-h(a)]/h (the algebraic manipulations to get from one to the other would not necessarily be valid)

22. Use the definition of the derivative twice, once as is and once with h replaced by -h. Average the two.

28a. Use f(x) = x^3 for x >= 0, -x^3 for x < 0. Then differentiate each normally, checking if both match at 0 to see if the derivative exists there
b. similar steps