# If  f(x) = ( x^3 )/( x^2 - 25 ), what are the points of inflection, concavity and critical points?

Nov 30, 2016

Critical numbers: $0$ and $\pm \sqrt{75}$

#### Explanation:

Concave down on $\left(- \infty , - 5\right)$ and on $\left(0 , 5\right)$

Concave up on $\left(- 5 , 0\right)$ and on $\left(5 , \infty\right)$

Inflection point $\left(0 , 0\right)$

$f ' \left(x\right) = \frac{{x}^{2} \left({x}^{2} - 75\right)}{{x}^{2} - 25} ^ 2$

$f ' \left(x\right)$ DNE at $x = \pm 5$ but those are not in the domain, so they are not critical.

$f ' \left(x\right) = 0$ at $x = 0$ and at $x + \pm \sqrt{75}$ which are in the domain, so they are all critical numbers.

$f ' ' \left(x\right) = \frac{50 x \left({x}^{2} + 75\right)}{{x}^{2} - 25} ^ 3$ could change sign at $0$ and at $\pm 5$

On $\left(- \infty , - 5\right)$, $f ' ' \left(x\right) < 0$, so $f$ is concave down.
On $\left(- 5 , 0\right)$, $f ' ' \left(x\right) > 0$, so $f$ is concave up.
On $\left(0 , 5\right)$, $f ' ' \left(x\right) < 0$, so $f$ is concave down.
On $\left(5 , \infty\right)$, $f ' ' \left(x\right) < 0$, so $f$ is concave up.

The concavity changes at $x = - 5$, $0$ and $5$. An inflection point is a point of the graph where concavity changes. Since $- 5$ and $5$ are not in the domain of $f$, there are no IPs there, but $f \left(0\right) = 0$, so $\left(0 , 0\right)$ is an Infle pt.