Question 534799
If it's really {{{f(x)=(x+3)/(x^2-16)=(x+3)/(x+4)(x-4)}}}
the points where it does not exist are x=-4 and x=4, and all other x are in the domain.
I agree that the range is (-infinity, infinity).
If that's really the function, it changes sign 3 times (at -4, -3 an 3), with vertical asymptotes at {{{x=-4}}} and {{{x=4}}}
and a zero at {{{x=-3}}}.
There is no minimum or maximum. The first derivative is always negative, so the function decreases continuously in each of the 3 regions where it exists, going from positive infinity at the right of vertical asymptote {{{x=-4}}}
down to negative infinity on the left of vertical asymptote {{{x=4}}}.
To hug those vertical asymptotes like that, the function has to go  from concave upwards to concave downwards. The second derivative must have an inflection point somewhere between them. I calculated it, but I may have made a mistake. I got {{{(2x^3+18x^2+96x+96)/(x^2-16)^3}}}
which is zero at a point near {{{x=-1.25}}}
You do not need both derivatives to be zero to have an inflection point.
If they are both zero, it is a saddle point, an inflection point where the slope is zero. An example of a function with an inflection point where the first derivative is not zero is {{{tan(x)}}} at {{{x=0}}}