# What are the critical values, if any, of f(x)= x^3/sqrt(x + 25)?

Jan 7, 2017

Critical points at $x = 0$

#### Explanation:

First, differentiate. Let $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$. Then $g \left(x\right) = {x}^{3}$ and $h \left(x\right) = \sqrt{x + 25}$.

We can differentiate $g \left(x\right)$ using the power rule. $g ' \left(x\right) = 3 {x}^{2}$

We need the chain rule for $h \left(x\right)$. Let $y = \sqrt{u} = {u}^{\frac{1}{2}}$ and $u = x + 25$. Then dy/(du) = 1/2u^(-1/2) = 1/(2u^(1/2). and $\frac{\mathrm{du}}{\mathrm{dx}} = 1$.

$h ' \left(x\right) = \frac{1}{2 {u}^{\frac{1}{2}}} \cdot 1 = \frac{1}{2 \sqrt{x + 25}}$

Use the quotient rule now.

$f ' \left(x\right) = \frac{g ' \left(x\right) \cdot h \left(x\right) - g \left(x\right) \cdot h ' \left(x\right)}{h \left(x\right)} ^ 2$

$f ' \left(x\right) = \frac{3 {x}^{2} \left(\sqrt{x + 25}\right) - {x}^{3} / \left(2 \sqrt{x + 25}\right)}{\sqrt{x + 25}} ^ 2$

$f ' \left(x\right) = \frac{\frac{6 {x}^{2} \left(x + 25\right) - {x}^{3}}{2 \sqrt{x + 25}}}{x + 25}$

$f ' \left(x\right) = \frac{5 {x}^{3} + 150 {x}^{2}}{2 {\left(x + 25\right)}^{\frac{3}{2}}}$

There will be critical points whenever the derivative equals $0$ or is undefined. The derivative is undefined at $x = - 25$. Set the derivative to $0$ to find the other critical points.

$0 = \frac{5 {x}^{3} + 150 {x}^{2}}{2 {\left(x + 25\right)}^{\frac{3}{2}}}$

$0 = 5 {x}^{3} + 150 {x}^{2}$

$0 = 5 {x}^{2} \left(x + 30\right)$

$x = 0 \mathmr{and} - 30$

So, there will be additional critical points at $x = 0$ and $x = - 30$.

However, to be a critical number, the function has to be defined at the given point. Therefore, $x = - 30$ and $x = - 25$ are not critical numbers. $x = 0$ is the only critical number.

Hopefully this helps!