# What are the points of inflection of f(x)= (x^2 - 8)/(x+3) ?

##### 1 Answer
Aug 11, 2017

There are no points of inflections

#### Explanation:

We need

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

Here,

$f \left(x\right) = \frac{{x}^{2} - 8}{x + 3}$

$u \left(x\right) = {x}^{2} - 8$, $\implies$, $u ' \left(x\right) = 2 x$

$v \left(x\right) = x + 3$, $\implies$, $v ' \left(x\right) = 1$

Therefore, the first derivative is

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

$= \frac{2 {x}^{2} + 6 x - {x}^{2} + 8}{x + 3} ^ 2$

$= \frac{{x}^{2} + 6 x + 8}{x + 3} ^ 2$

And the second derivative is

$u \left(x\right) = {x}^{2} + 6 x + 8$, $\implies$, $u ' \left(x\right) = 2 x + 6$

$v \left(x\right) = {\left(x + 3\right)}^{2}$, $\implies$, $v ' \left(x\right) = 2 \left(x + 3\right)$

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

$= \frac{\left(2 x + 6\right) \left(x + 3\right) - 2 \left({x}^{2} + 6 x + 8\right)}{x + 3} ^ 3$

$= \frac{\left(2 {x}^{2} + 6 x + 6 x + 18\right) - \left(2 {x}^{2} + 12 x + 16\right)}{x + 3} ^ 3$

$= \frac{2}{x + 3} ^ 3$

The inflection points are when $f ' ' \left(x\right) = 0$

$\frac{2}{x + 3} ^ 3 = 0$, $\implies$, $S = \emptyset$

graph{(x^2-8)/(x+3) [-58.53, 58.55, -29.24, 29.23]}