# How do you find the x coordinates of all points of inflection, find all discontinuities, and find the open intervals of concavity for y=x^5-2x^3?

May 22, 2017

Therefore, $y$ has points of inflection at $x = - \sqrt{\frac{3}{5}} , x = 0 , x = \sqrt{\frac{3}{5}}$, $y$ has no discontinuities, and is concave down for $x \in \left(- \infty , - \sqrt{\frac{3}{5}}\right) \cup \left(0 , \sqrt{\frac{3}{5}}\right)$, and concave up for $x \in \left(- \sqrt{\frac{3}{5}} , 0\right) \cup \left(\sqrt{\frac{3}{5}} , \infty\right)$.

#### Explanation:

$y = {x}^{5} - 2 {x}^{3}$

To find points of inflection and changes in concavity, find the second derivative of $y$, which is $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$, and set it equal to zero.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{4} - 6 {x}^{2}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 20 {x}^{3} - 12 x$

Factor:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 4 \left(x\right) \left(5 {x}^{2} - 3\right)$

Set $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ equal to zero, and use the zero product rule to evaluate:
$0 = 4 \left(x\right) \left(5 {x}^{2} - 3\right)$

$\textcolor{b l u e}{x = 0}$

$5 {x}^{2} - 3 = 0$
$5 {x}^{2} = 3$
${x}^{2} = \frac{3}{5}$

$\textcolor{b l u e}{x = \pm \sqrt{\frac{3}{5}}}$

Draw a sign line for the signs of $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$:

Therefore, $y$ has points of inflection at $x = - \sqrt{\frac{3}{5}} , x = 0 , x = \sqrt{\frac{3}{5}}$, $y$ has no discontinuities, and is concave down for $x \in \left(- \infty , - \sqrt{\frac{3}{5}}\right) \cup \left(0 , \sqrt{\frac{3}{5}}\right)$, and concave up for $x \in \left(- \sqrt{\frac{3}{5}} , 0\right) \cup \left(\sqrt{\frac{3}{5}} , \infty\right)$.