Question

How do you find the exact relative maximum and minimum of the polynomial function of `f(x)=x^3+6x^2-36x`?

Answer 1

The function has a relative minimum at `x=2`
The function has a relative maximum at `x=-6`

Explanation:

Given -

`y=x^3+6x^2-36x`

`dy/dx=3x^2+12x-36`
`(d^2y)/(dx^2)=6x+12`

`dy/dx=0 => 3x^2+12x-36=0`

`3x^2+12x-36=0` [Dividing both sides by 3 we get]
`x^2+4x-12=0`
`x^2+6x-2x-12=0`
`x(x+6)-2(x+6)=0`
`(x-2)(x+6)=0`

`x` has two values

`x-2=0`
`x=2`

`x+6=0`
`x=-6`

At `x=2`

`(d^2y)/(dx^2)=6(2)+12=12+12=24>0`

At `x=2`; `dy/dx=0`;`(d^2y)/(dx^2)>0`

Hence the function has a relative minimum at `x=2`

At `x=-6`

`(d^2y)/(dx^2)=6(-6)+12=-36+12=-24<0`

At `x=-6`; `dy/dx=0`;`(d^2y)/(dx^2)<0`

Hence the function has a relative maximum at `x=-6`

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