Question

Use the second derivative test to find the maximum and minimum points of f(x) = 2x^3+15x^2-36x. Help!? Help!?

Answer 1

`"maximum at " (-6,324)`
`"minimum at " (1,-19)`

Explanation:

We have:

`f(x) = 2x^3+15x^2-36x`

At a max/min the first derivative vanishes so we compute the first derivative:

`f'(x) = 6x^2+30x-36`

So at a max/min we require that:

`f'(x) = 0 => 6x^2+30x-36 = 0`
`:. x^2+5x-6 = 0`
`:. (x+6)(x-1) = 0`
`:. x=-6,1`

To determine the nature of the turning points we perform the second derivative test, and so we compute the second derivative:

`f''(x) = 12x+30`

And for the located critical points we have:

`x=-6 => f(x)=324, \ \ \ f''(x) lt 0`
`x=1 \ \ \ \ \ => f(x)=-19, f''(x) gt 0`

And so we can conclude that:

`"maximum at " (-6,324)`
`"minimum at " (1,-19)`

And we can verify these results graphically:
graph{2x^3+15x^2-36x [-15, 10, -200, 500]}