How do you find the critical numbers of #f(x)=ln(x^4+27)#?
1 Answer
Dec 7, 2016
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!