Question: Let f: be a function that is differentiable everywhere and such that f ’ is also differentiable in all of . Assume that there is a real number x ₀ such that f ’(x ₀ ) = 0 and that f’’(x) 0 for all x Here f"(x) denotes the derivative function of f ’ at x Prove that

f(x ₀ ) f(x) for all x

Such x ₀ is called a Global Minimum of the function f. Prove further that if f"(x) > 0 for all x (rather than f(x) 0), there can be at most one number x ₀ with f’ (x ₀ ) = 0.

Hint : Recall Corollary A which states

Let f be differentiable function on an interval (a, b). Then

1. f is strictly increasing if f ’(x) > 0 for all x ;
2. f is strictly decreasing if f ’(x) < 0 for all x ;
3. f is increasing if f ’(x) 0 for all x ;
4. f is decreasing if f ’(x) 0 for all x

What does the fact f"(x) 0 for all x say about f ’? and what does this , along with f ’(x ₀ ) = 0, say about f? How do things change if, instead of f"(x) 0, we have f"(x) > 0 for all x ??

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