showing a local minimum of a function I have a problem that is a little bit of a struggle for me. I'm pretty sure what I'm trying to prove is a minimum at the point a, but how I go about proving it is eluding me. Any help would be greatly appreciated!
Let $f'(a) = 0$ and $f''(a) > 0$. Show that there is a positive number $\delta$ such that $f(x) \ge f(a)$ for every $x$ with $|x - a| < \delta$.
 A: the Taylor theorem codifies some detailed computation. let $\Omega$ be an open interval containing $a$. then if for $x \in \Omega$ the derivatives $f^{(j)}(x)$ exist for $0 \le j \le n+1$ then $\exists b.|b-a| \le |x-a|$ and
$$
f(x) = \sum_{j=0}^n\frac1{j!}f^{(j)}(a)(x-a)^j + \frac1{(n+1)!}f^{n+1}(b)(x-a)^{n+1}
$$
in the present case we need $n=1$ giving
$$
f(x) = f(a) + \frac12 f''(b)(x-a)^2
$$
A: You have to use the Taylor expansion of the function at $a$!
A: This statement is called the second derivative test. Here's a simple proof.
By definition the second derivative is
$$
f''(a) = \lim_{h \to 0}{\frac {f'(a+h)-f'(a)}{h}}=\lim_{h \to 0}{\frac {f'(a+h)}{h}}
$$
Because $f''(a)>0$ and $f'(x)$ is differentiable and thus continuous, for very small $h$:
$$
\frac {f'(a+h)}{h} > 0 \Rightarrow \\
\Rightarrow
\begin{cases}
f'(x)<0 & \text{for} \ x<a \\
f'(x)>0 & \text{for} \ x>a
\end{cases}
$$
Having these two inequalities for $f'(x)$ near $a$ you can use the exact same reasoning as before on the definition of $f'(x)$.
$$
f'(a) = \lim_{h \to 0}{\frac {f(a+h)-f(a)}{h}} \Rightarrow \text{...} \Rightarrow f(a+h)>f(a)
$$
