# What are the points of inflection of f(x)=3x^5 - 5x^3 #?

Jan 4, 2016

They are
$\left(- \frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{4}\right) , \left(0 , 0\right) , \left(1 , - \frac{\sqrt{2}}{4}\right)$

#### Explanation:

Possible point of inflection is determine by setting the derivative equal to zero. Possible point on inflection happen when there is changes in concavity (refer to the graph below)

$f \left(x\right) = 3 {x}^{5} - 5 {x}^{3}$

$f ' \left(x\right) = 15 {x}^{4} - 15 {x}^{2}$

$f ' ' \left(x\right) = 60 {x}^{3} - 30 x$

Set the second derivative equal to zero

$60 {x}^{3} - 30 x = 0$

$30 x \left(2 {x}^{2} - 1\right) = 0$

$30 x = 0 \implies x = 0$

${x}^{2} - 1 = 0 \implies 2 {x}^{2} = 1$

${x}^{2} = \frac{1}{2} > x = \pm \sqrt{\frac{1}{2}} \implies \pm \frac{\sqrt{2}}{2}$

Coordinates point for P.O.I (Point of inflection)

$f \left(- \frac{\sqrt{2}}{2}\right) = 3 {\left(- \frac{\sqrt{2}}{2}\right)}^{5} - 5 \left(- \frac{\sqrt{2}}{2}\right)$

$\left(3\right) \left(- \frac{\sqrt{2}}{8}\right) - 5 \left(- \frac{\sqrt{2}}{8}\right) = - 3 \frac{\sqrt{2}}{8} + 5 \frac{\sqrt{2}}{8}$

$2 \frac{\sqrt{2}}{8} \implies \frac{\sqrt{2}}{4} = \implies \left(- \frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{4}\right)$

$f \left(0\right) = 0 - 0 = 0 \implies \left(0 , 0\right)$

$f \left(\frac{\sqrt{2}}{2}\right) = 3 {\left(\frac{\sqrt{2}}{2}\right)}^{5} - 5 \left(\frac{\sqrt{2}}{2}\right)$

$\left(3\right) \left(\frac{\sqrt{2}}{8}\right) - 5 \left(\frac{\sqrt{2}}{8}\right) = 3 \frac{\sqrt{2}}{8} - 5 \frac{\sqrt{2}}{8}$

$- 2 \frac{\sqrt{2}}{8} \implies - \sqrt{2} = \implies \left(- \frac{\sqrt{2}}{2} , - \frac{\sqrt{2}}{4}\right)$

graph{3x^5-5x^3 [-10, 10, -5, 5]}