# What is the x-coordinate of the point of inflection on the graph of y=1/10x^(5)+1/2X^(4)-3/10?

Jul 14, 2018

The x-coordinate of the inflexion point is $- 3$

#### Explanation:

$y = \frac{1}{10} {x}^{5} + \frac{1}{2} {x}^{4} - \frac{3}{10}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {x}^{4} + 2 {x}^{3}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2 {x}^{3} + 6 {x}^{2}$

For second derivative, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$

$2 {x}^{3} + 6 {x}^{2} = 0$
$2 {x}^{2} \left(x + 3\right) = 0$
$x = 0$ or $x = - 3$

Then you must these points for concavity
Test $\left(0 , - \frac{3}{10}\right)$
When $x = - 1$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 4$
When $x = 0$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$
When $x = 1$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 8$
Therefore, $\left(0 , - \frac{3}{10}\right)$ is not a point of inflexion as the concavity doesn't change

Test $\left(- 3 , 15 \frac{9}{10}\right)$,
When $x = - 4$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 32$
When $x = - 3$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0$
When $x = - 2$, $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 8$
Therefore, $\left(- 3 , 15 \frac{9}{10}\right)$ is a point of inflexion as the concavity changes