How do you find the local extremas for #f(x)=x^(1/3)(x+8)#?

1 Answer
May 16, 2016

Answer:

Has a local minimum at #x = -2#

Explanation:

The critical points are obtained by solving
#d/(dx)f(x)=(4 (2 + x))/(3 x^(2/3))=0#
Solving for #x# we get #x = -2#
The critical point qualification is done by calculating #d^2/(dx)^2 f(x) = (4 (-4 + x))/(9 x^(5/3))#
So #d^2/(dx)^2 f(-2) = 0.839947#
Then the critical point is a minimum.
If you where using a symbolic processor be aware of
#x^{1/3} equiv x/(abs x)abs x^{1/3}# and #x^{5/3} equiv x/(abs(x))abs[x]^(5/3)#