# How do you find the inflection points of the graph of the function: f(x) = 2x(x-4)^3?

Jun 6, 2015

The inflection points are given when the second derivative is equaled to zero, because the second derivative indicates the change in concavity of the function.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {\left(x - 4\right)}^{3} + 2 x \left(3 {\left(x - 4\right)}^{2}\right) = 2 {\left(x - 4\right)}^{3} + 6 x {\left(x - 4\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\left(2 x - 8\right) + \left(6 x\right)\right] \left[{\left(x - 4\right)}^{2}\right] = \textcolor{g r e e n}{\left(8 x - 8\right) {\left(x - 4\right)}^{2}}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 8 {\left(x - 4\right)}^{2} + \left(8 x - 8\right) \left(2 x - 8\right)$

Developing:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 8 \left({x}^{2} - 8 x + 16\right) + \left(16 {x}^{2} - 80 x + 64\right) = 24 {x}^{2} - 144 x + 192$

Equaling the second derivative to zero:

$24 \left({x}^{2} - 6 x + 8\right) = 0$

$\frac{6 \pm \sqrt{36 - \left(4\right) \left(1\right) \left(8\right)}}{2}$
$\frac{6 \pm \sqrt{4}}{2}$
$\frac{6 \pm 2}{2}$

${x}_{1} = 4$ and ${x}_{2} = 2$

Substituting these values found in the second derivative there in the original function, we'll get the inflection points.

$f \left(4\right) = 2 \left(4\right) {\left(4 - 4\right)}^{3} = 0$
$f \left(2\right) = 2 \left(2\right) {\left(2 - 4\right)}^{3} = 4 \cdot \left(- 8\right) = - 32$

Thus, the inflection points are $A \left(4 , 0\right)$ and $B \left(2 , - 32\right)$