Question

Use the Second Derivative Test to show that (3,27) is a maximum point on the graph of f(x) = x^3(4-x). Help!? Help!?

Answer 1

graph{x^3(4-x) [-10, 10, -10, 32.16]}

We have:

`f(x) = x^3(4-x)`
`" " = 4x^3 -x^4`

First check `(3,27)` lies on the curve:

`x = 3 => f(3) = 27(4-3) = 27`

Differentiate once to get the First Derivative:

`f'(x) = 12x^2 -4x^3`

At a min/max the first derivative is zero:

`f'(x) = 0 => 12x^2 -4x^3 = 0`
`:. 4x^2(3-x) = 0`
`:. x = 0,3`

So we have shown that '`x=3` is indeed a critical point. Now differentiate again to obtain the second derivative:

`f''(x) = 24x -12x^2`

For any critical point ()where `f'(x)=0`) then:

If `{ (f''(x) lt 0, =>, "maximum"), (f''(x) gt 0, =>, "minimum") :}`

So when `x=3` we have:

`f''(3) = 72 -108 lt 0 => "maximum"`

Hence, `(3,27)` lies on the curve and is a critical point corresponding to a local maximum. QED