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

##### 1 Answer

Dec 7, 2016

#### Answer:

#### Explanation:

Differentiate.

We let

#dy/dx = dy/(du) xx (du)/dx#

#dy/dx = 1/u xx 4x^3#

#dy/dx = (4x^3)/(x^4 + 27)#

The function will have critical numbers when

Let's start by finding any asymptotes.

#x^4 + 27 =0#

#x^4 = -27#

#x = root(4)(-27)#

Now, set the derivative to

#0 = (4x^3)/(x^4 + 27)#

#0 = 4x^3#

#x = 0#

Thus, the only critical point is at

Hopefully this helps!