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

1 Answer
Dec 26, 2015

Answer:

Minima #(0, 0)#
Maxima #(-4/3, 1 5/27)#

Explanation:

Given-

#y=x^2(x+2)#
#y=x^3+2x^2#
#dy/dx=3x^2+4x#
#(d^2y)/(dx^2)=6x+4#
#dy/dx=0=>3x^2+4x=0#
#x(3x+4)=0#
#x=0#
#3x+4=0#
#x=-4/3#
At #x=0; (d^2y)/(dx^2)=6(0)+4=4>0#

At #x=0; dy/dx=0;(d^2y)/(dx^2)>0#
Hence the function has a minima at #x=0#

At #x=0;y=(0)^2(0+2)=0#
Minima #(0, 0)#

At #x=-4/3; (d^2y)/(dx^2)=6(-4/3)+4=-4<0#
At #x=-4; dy/dx=0;(d^2y)/(dx^2)<0#

Hence the function has a maxima at #x=-4/3#

At #x=-4/3;y=(-4/3)^2(-4/3+2)=1 5/27#

Maxima #(-4/3, 1 5/27)#

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