Question

How do you use the Second Derivative Test where applicable to find the extrema for `f(x) = x^2(6-x)^3`?

Answer 1

See the explanation section below.

Explanation:

`f(x) = x^2(6-x)^3`

We need to fins the critical numbers for `f` first. Then we will use the second derivative test (where applicable) to test them.

`f'(x) = 2x(6-x)^3+x^2(3(6-x)^2 (-1))`

`= x(6-x)^2(12-5x)`

Critical numbers are `0`, `6`, and `12/5`.

We need the second derivative. Because we do not need the zeros of the second derivative, some people may prefer to expand the first derivative to write it in standard polynomial form. I want to keep factored form.

The product rule for three factors is
`(uvw)' = u'vw+uv'w+uvw'`

`f''(x) = (1)(6-x)^2(12-5x)+x2(6-x)(-1)(12-5x)+x(6-x)^2(-5)`

`= (6-x)[(6-6)(12-5x)-2x(12-5x)-5x(6-x)]`

`= 4(6-x)(5x^2-24x+18)`

Test the critical numbers:

`f''(0) = 4(6)(18)` is positive, so #f(0) is a local minimum.

`f''(6)=4(0)(2(6)^2-24(6)+18) = 0` so the second derivative test is not applicable. (The first derivative test shows that `f(6)` is not an extremum.

`f''(12/5) = f''(2.4) = 4(6-2.4)(2(2.4)^2-24(2.4)+18)`

The first two factors are clearly positive.

The third is
`2(2.4)(2.4)-24(2.4)+18 = 4.8(2.4)-24(2.4)+18`

`= -19.2(2.4)+18` which is negative. (`19.2 xx 2.4 > 18`)

So, `f''(12/5)` is negative, and `f(12/5)` is a local maximum.