# How do you find the points of Inflection of f(x)=2x(x-4)^3?

Apr 4, 2018

Below

#### Explanation:

$f \left(x\right) = 2 x {\left(x - 4\right)}^{3}$
$f ' \left(x\right) = 2 x \times 3 {\left(x - 4\right)}^{2} + 2 {\left(x - 4\right)}^{3}$
$f ' \left(x\right) = 2 {\left(x - 4\right)}^{2} \left(3 x + x - 4\right) = 2 {\left(x - 4\right)}^{2} \left(4 x - 4\right)$
$f ' ' \left(x\right) = 2 \times \left(2 \left(x - 4\right) \left(4 x - 4\right) + {\left(x - 4\right)}^{2} \left(4\right)\right)$
$f ' ' \left(x\right) = 2 \left(8 {x}^{2} - 40 x + 32 + 4 {x}^{2} - 32 x + 64\right)$
$f ' ' \left(x\right) = 2 \left(12 {x}^{2} - 72 x + 96\right)$
$f ' ' \left(x\right) = 2 \left(12\right) \left({x}^{2} - 6 x + 8\right)$
$f ' ' \left(x\right) = 24 \left(x - 4\right) \left(x - 2\right)$

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

ie
$24 \left(x - 4\right) \left(x - 2\right) = 0$
$x = 4 , 2$

Test $x = 4$
At $x = 3$, $f ' ' \left(x\right) = - 24$
At $x = 4$, $f ' ' \left(x\right) = 0$
At $x = 5$, $f ' ' \left(x\right) = 72$

Therefore, there is a change in concavity so there is a point of inflexion at x=4

Test $x = 2$
At $x = 1$, $f ' ' \left(x\right) = 72$
At $x = 2$, $f ' ' \left(x\right) = 0$
At $x = 3$, $f ' ' \left(x\right) = - 24$

Therefore there is a change in concavity so there is a point of inflexion at x=2

Apr 4, 2018

Point of inflection is at $x = 4$ or $x = \frac{8}{3}$

#### Explanation:

Points of Inflection appear where the curve changes from being concave to convex or vice versa. This happens when second derivative i.e. $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = 0$

As $f \left(x\right) = 2 x {\left(x - 4\right)}^{3}$

$\frac{\mathrm{df}}{\mathrm{dx}} = 2 {\left(x - 4\right)}^{3} + 2 x {\left(x - 4\right)}^{2} = 2 {x}^{3} - 24 {x}^{2} + 96 x - 128 + 2 {x}^{3} - 16 {x}^{2} + 32 x = 4 {x}^{3} - 40 {x}^{2} + 128 x - 128$

and $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = 12 {x}^{2} - 80 x + 128$

This is $0$, when $12 {x}^{2} - 80 x + 128 = 0$

or $3 {x}^{2} - 20 x + 32 = 0$

or $\left(x - 4\right) \left(3 x - 8\right) = 0$

Hence point of inflection is at $x = 4$ and $x = \frac{8}{3}$

Graph not drawn to scale.

graph{2x(x-4)^3 [-10, 10, -70, 30]}