Question

What are the absolute extrema of `f(x)=x^(1/3)*(20-x)in[0,20]`?

Answer 1

The absolute minimum is `0`, which occurs at `x = 0` and `x=20`.

The absolute maximum is `15root(3)5`, which occurs at `x = 5`.

Explanation:

The possible points that could be absolute extrema are:

1. Turning points; i.e. points where `dy/dx = 0`

2. The endpoints of the interval

We already have our endpoints (`0` and `20`), so let's find our turning points:

`f'(x) = 0`

`d/dx(x^(1/3)(20-x)) = 0`

`1/3x^(-2/3)(20-x) - x^(1/3) = 0`

`(20-x)/(3x^(2/3)) = x^(1/3)`

`(20-x)/(3x) = 1`

`20-x = 3x`

`20 = 4x`

`5 = x`

So there is a turning point where `x = 5`. This means that the 3 possible points that could be extrema are:

`x = 0" "" "x=5" "" "x=20`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let's plug these values into `f(x)`:

`f(0) = (0)^(1/3) (20 - 0) = 0 * 20 = color(red)0`

`f(5) = (5)^(1/3) (20 - 5) = root(3)(5) * 15 = color(red)(15root(3)5`

`f(20) = (20)^(1/3) (20-20) = root(3)(20) * 0 = color(red)0`

Therefore, on the interval `x in [0, 20]`:

The absolute minimum is `color(red)0`, which occurs at `x = 0` and `x=20`.

The absolute maximum is `color(red)(15root(3)5)`, which occurs at `x = 5`.

Final Answer