How to mathematically prove the optimality conditions for a univariate function? Consider a univariate function $f(x)$. I know the graphical intuition behind why $f'(x)=0$ at the extrema of $f$. But how do you prove it mathematically? 
I start with the assumption of $x^*$ being a minimum (the maximum case can be proved likewise), then
$f(x^*+h) \geq f(x^*)$ where $h\in \mathcal{N}(x)$, $\mathcal{N}$ being the neighborhood. This gives,
$f(x^*) + hf'(x^*) + \frac{h^2}{2!} f''(x^*) + \ldots \geq f(x^*)$.
How do I proceed from here to prove that $f'(x^*) = 0$ and $f''(x^*) \geq 0$ ? This seems so simple but is eluding me.
NOTE: I do not want the proof for $\nabla f(\mathbf{x^*}) = 0$ and $\nabla^2 f(\mathbf{x^*}) \succeq 0$ for multivariate function.

When $f$ is differentiable at  $a$  we can write
$$f(x)-f(a)=m(x)\>(x-a)\qquad\bigl(x\in{\rm dom}(f)\bigr)\ ,\tag{1}$$
whereby the function $$x\mapsto m(x):=\cases{{f(x)-f(a)\over \mathstrut x-a}\quad&$(x\ne a)$\cr  f'(a)&$(x=a)$\cr}$$
is continuous at $a$. If $f'(a)\ne0$ then $${\rm sgn}\bigl(m(x)\bigr)={\rm sgn}\bigl(f'(a)\bigr)\ne0$$ for all $x$  in a full neighborhood of $a$. From $(1)$ it then follows that $f(x)-f(a)$ takes as well positive as negative values in the immediate neighborhood of $a$. Under such circumstances $f$ cannot have a local extremum at $a$.
From this we conclude that $f'(a)=0$ is a necessary condition for a local extremum at $a$.
