Question

How do you find the relative extrema for `f(x) = x^2(6-x)^3`?

Answer 1

Minimum when `x=0`, maximum when `x=12/5`

Explanation:

Find the critical values of the function. These occur when the derivative equals `0` or is undefined.

To find the derivative of the function, first find the derivative. Although you could distribute the equation, it's probably easier to use the product rule.

`f'(x)=(6-x)^3d/dx[x^2]+x^2d/dx[(6-x)^3]`

Find each derivative (the second requires the chain rule):

`d/dx[x^2]=2x`

`d/dx[(6-x)^3]=3(6-x)^2*d/dx[6-x]=-3(6-x)^2`

Plug these back in.

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

Factor `x(6-x)^2` from each term.

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

Simplify.

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

This is never undefined. It is equal to `0` when `x=0`, `x=6`, or `x=12/5`.

We can determine what types of extrema these are using the first derivative test (see how the signs change around the points).

Determining `x=0`:

When `x<0`, `f'(x)<0`.
When `0 < x <12/5,f'(x)>0`
Since the derivative changes from decreasing to increasing, there is a minimum at `x=0`.

Determining `x=12/5`:

When `0 < x <12/5,f'(x)>0`
When `12/5 < x < 6,f'(x)<0`
Since the derivative changes from increasing to decreasing, there is a maximum at `x=12/5`.

Determining `x=6`:

When `12/5 < x < 6,f'(x)<0`
When `x>6,f'(x)<0`
Here, the sign of the second derivative doesn't change. This means it is not an extremum, just a point where graph temporarily flattens.

graph{x^2(6-x)^3 [-3, 9, -100, 300]}