# What are the points of inflection, if any, of f(x) = (x+1)^9-9x-2 ?

Jan 13, 2016

It has one point of infection at $\left(- 1 , 7\right)$

#### Explanation:

Points of inflection occur when the curve changes from concave up to concave down or vice versa. If the second derivative is positive then the function is concave up. If the second derivative is negative then the function is concave down. If the second derivative is continuous and changes sign, then it will be zero at the point where it changes sign.

Given:

$f \left(x\right) = {\left(x + 1\right)}^{9} - 9 x - 2$

Differentiate using the power and chain rules to find:

$f ' \left(x\right) = 9 {\left(x + 1\right)}^{8} - 9$

$f ' ' \left(x\right) = 72 {\left(x + 1\right)}^{7}$

So $f ' ' \left(x\right) = 0$ when $x + 1 = 0$, that is when $x = - 1$

Note that if $x < - 1$ then $f ' ' \left(x\right) < 0$ and if $x > - 1$ then $f ' ' \left(x\right) > 0$, so this is definitely a point of inflection.

$f \left(- 1\right) = 7$

So the point of inflection is at $\left(- 1 , 7\right)$