Show that if the curve $y = f(x)$ has a maximum stationary point at $x = a$ I am looking for help in solving the following problem:

Show that if the curve $y = f(x)$ has a maximum stationary point at $x = a$ then the curve $y = \frac {1}{f(x)}$ has a minimum stationary point at $x = a$ as long as $f(a) \neq 0$.

Thank you.
Edit: Thank you for all the comments. However, I made a mistake in my post, which changes the exercise, and the answer, obviously. I wrongly typed =, where it should be $\neq$. I would be glad if somebody helped me - I still do not know how to start.
 A: You could think about a specific example such as;
$$f(x)=\sin(x)$$
and its maximum at 
$$\big(\frac{\pi}{2},1\big)$$
and then about
$$\frac{1}{f(x)}=\csc(x)=\frac{1}{\sin(x)}$$
and it's minimum at
$$\big(\frac{\pi}{2},1\big)$$
to get a feel for the question before trying to generalise.
Or some other specific example.
(I've been trying to think of another, easier, example)
How about,
$$f(x)=1-x^2$$
with a maximum at (0,1)
and
$$\frac{1}{f(x)}=\frac{1}{1-x^2}$$
with a minimum at (0,1) ?
A: We have $$\left[\frac{1}{f(x)}\right]'=-\frac{f'(x)}{f(x)^2}$$ and $$\left[\frac{1}{f(x)}\right]''=-\left[\frac{f'(x)}{f(x)^2}\right]'=-\frac{f''(x)f(x)^2-2f(x)f'(x)^2}{f(x)^4}.$$ Then, at the point in question, being $x=a,$ by hypothesis we have $f(x)\neq 0, f'(x)=0,f''(x)<0.$ 
Can you complete it now?

As @TonyK has pointed out, this approach does not work, so I guess we'll have to go back to definitions.
If $f(x)$ has a maximum value at $x=a,$ then for all sufficiently small $\epsilon>0$ we must have $$\begin {align} {f'(a-\epsilon)>0 \\f'(a+\epsilon)<0}\end {align}.$$ A similar definition holds for minimum points, only now with the inequality symbols reversed. We now do not need second derivatives. Using first derivatives and the above definitions gives your result (together with the other assumptions, of course).
A: As Allawonder shows with 
$$\left[\frac{1}{f(x)}\right]'=-\frac{f'(x)}{f(x)^2}$$
$1/f$ has a stationary point at $a$. But to show that this point is a minimum, we need to go back to first principles:
$f(a)$ is a maximum, so there exists $\delta>0$ such that when $|x-a|<\delta$, then (i) $f(x)\le f(a)$ and (ii) $f(x)$ and $f(a)$ have the same sign.
But if $f(x)\le f(a)$, and $f(x)$ and $f(a)$ have the same sign, then $1/f(x)\ge 1/f(a)$. So $1/f(x)$ has a minimum at $a$.
