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

1 Answer
Dec 7, 2016

Answer:

#x = 0# is the only critical number.

Explanation:

Differentiate.

We let #y = lnu# and #u = x^4 + 27#. Then #dy/(du) = 1/u# and #(du)/dx = 4x^3#.

#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 #f'(x) = 0# or when #f'(x)# is undefined.

Let's start by finding any asymptotes.

#x^4 + 27 =0#

#x^4 = -27#

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

#:.#The function is defined for all real values of #x#.

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

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

#0 = 4x^3#

#x = 0#

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

Hopefully this helps!