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What are the local extrema, if any, of #f(x) =x^2(x+2) #?

1 Answer

Nov 7, 2017

#x=0, -4/3#

Explanation:

Find the derivative of #f(x)=x^2(x+2)#.
You will have to use the product rule.

#f'(x)=x^2+(x+2)2x=x^2+2x^2+4x=3x^2+4x#
#f'(x)=x(3x+4)#

Set #f'(x)# equal to zero to find the critical points.

#x=0#
#3x+4=0 rarr x=-4/3#

#f(x)# has local extrema at #x=0, -4/3#.
OR
#f(x)# has local extrema at the points (0, 0) and (#-4/3#, #32/27#).

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