Show that if the curve y=f(x) has a maximum stationary point at x=a, then the curve y=1/f(x) has a minimum stationary point at x=a as long as f(a) does not equal 0.
Stationary point is when the derivative of the function=0.

Question

mathmale · Answer

"Stationary point" is not a commonly seen vocabulary word in calculus, so you might want to  (1) look it up yourself and (2) share its definition with the rest of us.

anonymous · Answer

Stationary point is when the derivative of the function=0. Sorry for not clarifying

irishboy123 · Answer

@chau88  does have a point

anonymous · Answer

@SolomonZelman, Thank you very much for your answer. I believe that I missed out the numerator f'(x) when I was differentiating y. However, I don't really understand why 1/max=min. 
What I tried to do was second differentiate it and show that f''(x)>0 and therefore the function is max. But I realised that it was too hard to differentiate $$y'=f'(x)/f(x)^{2}$$
Please advise. Thank!

loser66 · Answer

Let $g(x) = \dfrac{1}{f(x)}$ 
We need show
1) a is a critical point of g(x)
2) g(a) is minimum. 
1) a is a critical point of g(x) :
 At x =a, $f'(a) =0$ and we know that f(a) max at a. 
Now, $g'(x) = \dfrac{-f'(x)}{f^2(x)}\implies g'(a) = \dfrac{-f'(a)}{f^2(a)}$
We have $f(a) \neq 0$, hence g'(a) defined
Moreover, f'(a) =0, Hence g'(a) =0. That shows a is a critical point of g(x).  
2) g(a) is minimum of g(x) 
Look at the chart for f(x) |dw:1450966314014:dw|