# What are the points of inflection of f(x)=x^5 - 2x^3 +4x?

Apr 21, 2017

The points of inflection of $f \left(x\right)$ are at
$x = - \sqrt{\frac{3}{10}} , x = 0 ,$ and $x = \sqrt{\frac{3}{10}}$

#### Explanation:

Use power rule to find $f ' ' \left(x\right)$

$f \left(x\right) = {x}^{5} - 2 {x}^{3} + 4 x$
$f ' \left(x\right) = 5 {x}^{4} - 6 {x}^{2} + 4$
$f ' ' \left(x\right) = 20 {x}^{3} - 6 x$

For points of inflection, set second derivative equal to zero:
$0 = f ' ' \left(x\right)$
$0 = 2 \left(x\right) \left(10 {x}^{2} - 3\right)$

Using zero product rule :
color(blue)(x=0

$10 {x}^{2} - 3 = 0$
${x}^{2} = \frac{3}{10}$
color(blue)(x=+-sqrt(3/10)

Double check that these answers are actual points of inflection by testing values in between these intervals (or draw a "sign line", as some call it).

Thus, the points of inflection of $f \left(x\right)$ are at
$x = - \sqrt{\frac{3}{10}} , x = 0 ,$ and $x = \sqrt{\frac{3}{10}}$