# How do you find the inflection points for the function f(x)=x^3+x?

Oct 10, 2014

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

By taking derivatives,

$f ' \left(x\right) = 3 {x}^{2} + 1$

$f ' ' \left(x\right) = 6 x = 0 R i g h t a r r o w x = 0$,

which is the $x$-coordinate of a possible inflection point. (We still need to verify that $f$ changes its concavity there.)

Use $x = 0$ to split $\left(- \infty , \setminus \infty\right)$ into $\left(- \infty , 0\right)$ and $\left(0 , \infty\right)$.

Let us check the signs of $f ' '$ at sample points $x = - 1$ and $x = 1$ for the intervals, respectively.
(You may use any number on those intervals as sample points.)

$f ' ' \left(- 1\right) = - 6 < 0 R i g h t a r r o w f$ is concave downward on $\left(- \infty , 0\right)$

$f ' ' \left(1\right) = 6 > 0 R i g h t a r r o w f$ is concave upward on $\left(0 , \infty\right)$

Since the above indicates that $f$ changes its concavity at $x = 0$, $\left(0 , f \left(0\right)\right) = \left(0 , 0\right)$ is an inflection point of $f$.

I hope that this was helpful.