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

Nov 26, 2017

There is no point of inflexion of $f \left(x\right)$ but there is a minimum point $\left(0 , - \frac{8}{3}\right)$.

#### Explanation:

Some basic rules of differentiation are as follows, where $u$ and $v$ are functions of $x$:

If $y = u \pm v$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(u\right) \pm \frac{d}{\mathrm{dx}} \left(v\right)$

2. Chain rule
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

3. Product rule
If $y = u v$, $\frac{\mathrm{dy}}{\mathrm{dx}} = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

4. Quotient rule
If $y = \frac{u}{v}$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

Let's get started,

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

First, differentiate $f \left(x\right)$,

$f ' \left(x\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left({x}^{2} - 8\right)\right] \left({x}^{2} + 3\right) - \left[\frac{d}{\mathrm{dx}} \left({x}^{2} + 3\right)\right] \left({x}^{2} - 8\right)}{{x}^{2} + 3} ^ 2$ ( Quotient rule )
$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{\left(2 x\right) \left({x}^{2} + 3\right) - \left(2 x\right) \left({x}^{2} - 8\right)}{{x}^{2} + 3} ^ 2$
$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{\left(2 x\right) \left(\cancel{{x}^{2}} + 3 \cancel{- {x}^{2}} + 8\right)}{{x}^{2} + 3} ^ 2$
$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{\left(2 x\right) \left(11\right)}{{x}^{2} + 3} ^ 2$
$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{22 x}{{x}^{2} + 3} ^ 2$

Since, it is an stationary point, $\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right] = 0$,

$\frac{22 x}{{x}^{2} + 3} ^ 2 = 0$
$\textcolor{w h i t e}{\frac{\times}{\times} x .} 22 x = 0$
$\textcolor{w h i t e}{\times \times \times} x = 0$

Find the $y$-coordinate of the stationary point by substituting $x = 0$ into $f \left(x\right)$,

$y = \frac{{\left(0\right)}^{2} - 8}{{\left(0\right)}^{2} + 3}$
$y = - \frac{8}{3}$

Hence, the co-ordinate of the stationary point is $\left(0 , - \frac{8}{3}\right)$.

To find the nature of the stationary point, differentiate $f ' \left(x\right)$,

$f ' ' \left(x\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(22 x\right)\right] {\left({x}^{2} + 3\right)}^{2} - \left[\frac{d}{\mathrm{dx}} {\left({x}^{2} + 3\right)}^{2}\right] \left(22 x\right)}{{x}^{2} + 3} ^ 4$ ( Quotient rule )
$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{\left(22\right) {\left({x}^{2} + 3\right)}^{2} - \left(4 x\right) \left({x}^{2} + 3\right) \left(22 x\right)}{{x}^{2} + 3} ^ 4$ ( Product rule )
$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{\left(22\right) \left({x}^{2} + 3\right) \left[\left({x}^{2} + 3\right) - \left(4 x\right) \left(x\right)\right]}{{x}^{2} + 3} ^ 4$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{\left(22\right) \left(3 - 3 x\right)}{{x}^{2} + 3} ^ 3$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{66 - 66 {x}^{2}}{{x}^{2} + 3} ^ 3$

Substitute $x = 0$ into $f ' ' \left(x\right)$,

$f ' ' \left(x\right) = \frac{66 - 66 {\left(0\right)}^{2}}{{\left(0\right)}^{2} + 3} ^ 3$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{66}{27} > 0$ ( minimum point )

Hence, the stationary point $\left(0 , - \frac{8}{3}\right)$ is, in fact, a minimum point and not an inflexion point.

Check:
graph{(x^2 - 8)/(x^2+3) [-10, 10, -5, 5]}