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. 

Let $g(x)=y=5/(3+f(x))$. For any $x$, we have
$$
3+f(x)\geq3+f(3)=3+2=5>0\implies g(x)=\frac{5}{3+f(x)}\leq\frac{5}{3+f(3)}=g(3).
$$
And so $y=g(x)$ is everywhere defined (because $3+f(x)>0$ always; see above) and has a global maximum at $x=3$.
