Suppose that f(x) and g(x) are positive functions such that f(x) has a local maximum and g(x) has a local minimum at x=c. Prove that

$\displaystyle h(x) = \frac{f(x)}{g(x)}$

this means if $\displaystyle f'(c) = 0$ and $\displaystyle g'(c) = 0$ that $\displaystyle h'(c) = 0$

$\displaystyle \frac{f'(x)g(x) - g'(x)f(x)}{[g'(x)]^2}$ at x = c $\displaystyle \Rightarrow \frac{0g(x) - 0f(x)}{0^2} = \frac{0}{0} = 0$

or

$\displaystyle \frac{f'(c)}{g'(c)} = h'(c) \Rightarrow \frac{0}{0} = 0$

I don't know if I am close.