# What are the points of inflection of f(x)=(3x^2 + 8x + 5)/(4-x) ?

Nov 27, 2017

Even though there are no inflexion points, there still are minimum$\left(- 1.323 , - 0.063\right)$ and maximum$\left(9.323 , - 63.937\right)$.

#### 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,

$f \left(x\right) = \frac{3 {x}^{2} + 8 x + 5}{4 - x}$

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

$f ' \left(x\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 8 x + 5\right)\right] \left(4 - x\right) - \left[\frac{d}{\mathrm{dx}} \left(4 - x\right)\right] \left(3 {x}^{2} + 8 x + 5\right)}{4 - x} ^ 2$
$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{\left(6 x + 8\right) \left(4 - x\right) - \left(- 1\right) \left(3 {x}^{2} + 8 x + 5\right)}{4 - x} ^ 2$
$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{- 6 {x}^{2} + 16 x + 32 + 3 {x}^{2} + 8 x + 5}{4 - x} ^ 2$
$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{3 {x}^{2} - 24 x - 37}{x - 4} ^ 2$

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

$- \frac{3 {x}^{2} - 24 x - 37}{x - 4} ^ 2 = 0$
$- 3 {x}^{2} + 24 x + 37 = 0$
$\textcolor{w h i t e}{\times \times \frac{\times}{\times} \times \times} x = - 1.32291 \mathmr{and} 9.32291$

Find the $y$-coordinate of the stationary points by substituting $x = - 1.32291 \mathmr{and} 9.32291$ into $f \left(x\right)$,

$y = - 0.063 \mathmr{and} - 63.937$

Hence, the co-ordinates of the stationary points are $\left(- 1.323 , - 0.063\right)$ and $\left(9.323 , - 63.937\right)$.

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

$f ' ' \left(x\right) = - \frac{\left[\frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 24 x - 37\right)\right] {\left(x - 4\right)}^{2} - \left[\frac{d}{\mathrm{dx}} {\left(x - 4\right)}^{2}\right] \left(3 {x}^{2} - 24 x - 37\right)}{x - 4} ^ 4$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = - \frac{\left(6 x - 24\right) {\left(x - 4\right)}^{2} - \left(2\right) \left(x - 4\right) \left(1\right) \left(3 {x}^{2} - 24 x - 37\right)}{x - 4} ^ 4$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = - \frac{\left(x - 4\right) \left[\left(6 x - 24\right) \left(x - 4\right) - \left(2\right) \left(1\right) \left(3 {x}^{2} - 24 x - 37\right)\right]}{x - 4} ^ 4$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = \frac{6 {x}^{2} - 48 x + 96 - 6 {x}^{2} + 48 x + 74}{x - 4} ^ 3$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = - \frac{170}{x - 4} ^ 3$

Let $x = - 1.32291$,

$f ' ' \left(x\right) = - \frac{170}{1.32291 - 4} ^ 3$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = 8.86054 > 0$ ( minimum )

Let $x = 9.32291$,

$f ' ' \left(x\right) = - \frac{170}{9.32291 - 4} ^ 3$
$\textcolor{w h i t e}{f ' ' \left(x\right)} = - 1.12720 < 0$ ( maximum )

Hence, minimum$\left(- 1.323 , - 0.063\right)$ and maximum$\left(9.323 , - 63.937\right)$.

Check:
graph{(3x^2 + 8x + 5)/(4-x) [-320, 320, -160, 160]}