# What are the points of inflection, if any, of f(x) =x/(x+8)^2?

Apr 4, 2018

Below

#### Explanation:

$f \left(x\right) = \frac{x}{x + 8} ^ 2$
$f ' \left(x\right) = \frac{{\left(x + 8\right)}^{2} \left(1\right) - \left(x\right) 2 \left(x + 8\right)}{x + 8} ^ 4$
$f ' \left(x\right) = \frac{\left(x + 8\right) - 2 x}{x + 8} ^ 3$
$f ' \left(x\right) = \frac{8 - x}{x + 8} ^ 3$
$f ' ' \left(x\right) = \frac{{\left(x + 8\right)}^{3} \left(- 1\right) - \left(8 - x\right) \left(3\right) {\left(x + 8\right)}^{2}}{x + 8} ^ 6$
$f ' ' \left(x\right) = \frac{\left(- 1\right) \left(x + 8\right) - 3 \left(8 - x\right)}{x + 8} ^ 4$
$f ' ' \left(x\right) = \frac{- x - 8 - 24 + 3 x}{x + 8} ^ 4$
$f ' ' \left(x\right) = \frac{2 x - 32}{x + 8} ^ 4$

For points of inflexion, $f ' ' \left(x\right) = 0$

ie

$\frac{2 x - 32}{x + 8} ^ 4 = 0$
$2 x - 32 = 0$
$x = 16$

Test $x = 16$

When:
$x = 15$, $f ' ' \left(x\right) = - \frac{2}{279841}$
$x = 16$, $f ' ' \left(x\right) = 0$
$x = 17$, $f ' ' \left(x\right) = \frac{2}{390625}$

Since there is a change in concavity, then there is a point of inflexion at $\left(16 , \frac{1}{36}\right)$