# How do you find the critical numbers of f(x)=ln(x^4+27)?

Dec 7, 2016

$x = 0$ is the only critical number.

#### Explanation:

Differentiate.

We let $y = \ln u$ and $u = {x}^{4} + 27$. Then $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 4 {x}^{3}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \times 4 {x}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {x}^{3}}{{x}^{4} + 27}$

The function will have critical numbers when $f ' \left(x\right) = 0$ or when $f ' \left(x\right)$ is undefined.

Let's start by finding any asymptotes.

${x}^{4} + 27 = 0$

${x}^{4} = - 27$

$x = \sqrt[4]{- 27}$

$\therefore$The function is defined for all real values of $x$.

Now, set the derivative to $0$ and solve.

$0 = \frac{4 {x}^{3}}{{x}^{4} + 27}$

$0 = 4 {x}^{3}$

$x = 0$

Thus, the only critical point is at $x = 0$.

Hopefully this helps!