# How do you find the extrema and points of inflection for f(x) =(3x)/((x+8)^2)?

$\left(\frac{3 x}{x + 8} ^ 2\right) ' = \frac{3 {\left(x + 8\right)}^{2} - 3 x \left(2 \left(x + 8\right)\right)}{x + 8} ^ 4 = - \frac{3 \left(x - 8\right)}{x + 8} ^ 3$
Root is $x = 8$ We have global maxima at $x = 8$ and $y = \frac{3}{32}$
Derivate again $- \frac{3 \left(x - 8\right)}{x + 8} ^ 3$ = $- \frac{3 {\left(x + 8\right)}^{3} - 9 \left(x - 8\right) {\left(x + 8\right)}^{2}}{x + 8} ^ 6 = \frac{6 \left(x - 16\right)}{x + 8} ^ 4$
Root is $x = 16$ We have a points of inflection at $x = 16$ and $y = \frac{1}{12}$