Question

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

Answer 1

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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