Question

If f(x) = ln(x^2 + 9)
a) find the local maximum and minimum values
I found the derivative and set it equal to 0 but idk what to do next.
b) find the interval of concavity and the inflection points

Answer 1

f' = 2x/(x^2+9)
Now, you know if a fraction p/q has its numerator p=0, then p/q=0
So, 2x/(x^2+9) = 0 when 2x=0. That is, when x=0, as we found above.

Also, since f" = -2(x^2-9)/(x^2+9)^2 you know that f"(x) = 0 at x = -3,3
so there are inflection points at x = -3,3
f"(x) > 0 for |x| < 3
f"(x) < 0 for |x| > 3
and since f"(0) > 0, f(x) has a minimum at (0,ln9)