If $f(x)$ has a minimum at (3,2) what does $y = \frac{5}{3+f(x)}$ have at x = 3? No calculus allowed. 
I can get the value by substituting 2 for $f(3)$, and that it should be a maximum turning point as $f(x+\delta h)$ and $f(x -\delta h) > f(x)$ but am not sure how to proceed further. 

Hint: $$f(3)=2, f(3+\varepsilon)>2; f(3-\varepsilon)>2$$
$$y=\frac{5}{3+f(x)}$$
$$y(3)=\frac5{3+2}=1$$
$$y(3+\varepsilon)<\frac5{3+2}=1$$
$$y(3-\varepsilon)<\frac5{3+2}=1$$
Then $g(x)$ has a maximum in $(3;1)$
