Is the opposite of the Second Derivative Test also true? Given the Second Derivative Test, one case says :
If $f(x_0)''<0$, then $f$ has a local maximum at $x_0$. 
Is it also true that, if $f$ has a local maximum at $x_0$,  $f(x_0)'' < 0$ ?
 A: As others have stated, the converse is not true (consider $f(x) = -x^4$ at $x_0=0$). 
However, one can still prove $f''(x_0)\le 0$ if $f$ is twice continuously differentiable:
Take $x$ close to $x_0$ then
$$
f(x_0) \ge f(x)  = f(x_0) + f'(x_0) +\frac12 f''(x_0)(x-x_0)^2 + o(|x-x_0|^2).
$$
Since $x_0$ is a local maximum, $f'(x_0)=0$. Then dividing the inequality by $(x-x_0)^2$ gives
$$
\frac 12f''(x_0) \le  r(x,x_0)
$$
with $r\to0$ if $x\to x_0$. Let now $x$ tend to zero, then $f''(x_0)\le 0$ follows.
If one knows in addition that the function $f$ behaves like a quadratic near $x_0$, i.e. there exists $\alpha>0$ such that
$$
f(x) + \frac\alpha2 |x - x_0|^2 \le f(x_0) 
$$
for all $x$ in a neighborhood of $x_0$, 
then  it follows as above $f''(x_0)\le -\alpha$.
A: First, note that the first statement is not the second derivative test - you also require that $f'(x_0) = 0$ which you do not write.
Second, the converse is not true. A local maximum may exist at a non-differentiable or non-twice differentiable point, for instance. 
If we further assume that $f'(x_0) = 0$ and $f$ is twice differentiable, the statement is still not true since you may have $f''(x_0) = 0$. For instance, take $f(x) = -x^4$. It has a local maximum at $x = 0$ but $f''(0) = 0$. Or we could take a trivial example of $f = 0$, too.
A: The converse is not true. Take $f(x)=x$ on $[0,1]$. It has a local maximum at $x=1$ but $f''(x)=0$.
A: This is a common misconception. Maybe the best way to avoid it is to understand why the original statement is true. 
If $f(x)$ is smooth enough at a critical point $x_0$, then we have
$$f(x)-f(x_0)=\frac12f''(x_0)(x-x_0)^2+o((x-x_0)^2)\quad \text{as}\quad x\to x_0$$
By $\epsilon-\delta$ definition it is easy to show that if $f''(x_0)>0$ then $x_0$ is a local minimum. 
But what about the converse? No!
When $f''(x_0)=0$, you won't know anything about the sign of $f(x)-f(x_0)$  if only according to the information of the second order derivative because it is $o((x-x_0)^2)$ as $x\to x_0$!
Yet, that is not to say you cannot know it anyway! Because you are still able to expand the Taylor series to a higher order. Say, provided that $f(x)$ is $(k+1)$-th differentiable at $x_0$, even if $f'(x_0)=f''(x_0)=\cdots=f^{(k)}(x_0)=0$ but $f^{(k+1)}(x_0)>0$ with $k$ an odd integer, it is still true that $x_0$ is a local minimum. 
