Explain why or why not:

Question

1. if f'(c)=0, then f has a local maximum or minimum at c.
2. if f''(c)=0, then f has an inflection point at c.
3. F'(x)=x^2+10 and G(x)=x^2-100 are antiderivatives of the same function
4. Between two local minima of a function continuous on (-∞,∞), there must be a local minimum.

Answer 1

I mistyped #4, it should be:
Between two local minima of a function continuous on (-∞,∞), there must be a local MAXIMUM.

Answer 2

#1. only if f" ≠ 0. Consider f(x) = x^3
#2. only if f' ≠ 0. Consider f(x) = x^4
#3. I guess so, if F" = G'
If you meant to type F(x) instead of F'(x), then I'd have to say true.
#4. f has a local min when f' changes from negative to positive.
So, if f(x) has minima at x=a,b with a < b, then for some some small h,
f'(a-h) < 0
f'(a+h) > 0
f'(b-h) < 0
f'(b+h) > 0
So, at some point c such that a+h < c < b-h we must have f'(c) = 0 and thus f"(c) < 0, meaning f(c) is a local max.
But what if f' is not continuous? It could switch from + to - without having to be 0 at x=c.
So, what about
f(x) =
x - x^2/2 + 3/2 for x <= 3/2
2x - x^2/2 for x > 3/2
the limit as x→3/2 is 3, so f(x) is continuous, but f'(x) is discontinuous at x = 3/2. f(x) as a minimum at both x=1,2 but no maximum in between.

If you strengthen the condition to inckude that both f and f' are continuous, then the statement is true.