Question

Let f be a differentiable function on [a,b ] and x belongs to[a,b]. Show that, if f'(x)=0 and f''(x)>0, then f must have a local maximum at x.

Answer 1

To prove: If $f'(x)=0$ and $f''(x)>0$ then $(x,f(x))$ is local minimum of $f$

Proof:

If a function f is continuous on the interval [a, b] and $x\in (a,b)$ and f(x)>b where b is some constant real number , then for all c near x, f(c)>b

Assume this theorem were not true. Then the intermediate value theorem would not hold

and similarly for the case f(x)<a

Let f be a function on [a, b] and let $f''(x\in (a,b))>0$ Then by above statement it must be that for some open interval around x containing x, f''(x)>0

If f''(x)>0 for some interval around x containing x, and f'(x)=0, then for all c in the interval,

$c<x\Rightarrow f'(c)<0\\c>x\Rightarrow f'(c)>0$

because f'(x) is an increasing function in the interval

We now have that x is a local minimum by the definition of local minimum and can also be considered the variant for x being a local maximum.