# How do you find the inflection points for f(x)=x^4-10x^3+24x^2+3x+5?

May 18, 2015

Inflection points are points of the graph of $f$ at which the concavity changes.
In order to investigate concavity, we look at the sign of the second derivative:

$f \left(x\right) = {x}^{4} - 10 {x}^{3} + 24 {x}^{2} + 3 x + 5$

$f ' \left(x\right) = 4 {x}^{3} - 30 {x}^{2} + 48 x + 3$

$f \left(x\right) = 12 {x}^{2} - 60 x + 48 = 12 \left({x}^{2} - 5 x + 4\right) = 12 \left(x - 1\right) \left(x - 4\right)$

So, $f ' '$ never fails to exist, and $f ' ' \left(x\right) = 0$ at $x = 1 , 4$

Consider the intervals:

$\left(- \infty , 1\right)$, $f ' ' \left(x\right)$ is positive, so $f$ is concave up
$\left(1 , 4\right)$, $f ' ' \left(x\right)$ is negative, so $f$ is concave down
$\left(4 , \infty\right)$, $f ' ' \left(x\right)$ is positive, so $f$ is concave up

The concavity changes at $x = 1$ and $y = f \left(1\right) = 23$.

The concavity changed again at $x = 4$ and $y = f \left(4\right) = 17$.

The inflection points are:

$\left(1 , 23\right)$ and $\left(4 , 17\right)$.