Question

Let `p(x)=x^3-9x^2+10` Find the absolute maximum and minimum value of `p` on the interval `[-2,8]`?

Answer 1

Absolute maximum: `(0, 10)`
Absolute minimum: `(6, -98)`

Explanation:

We start by differentiating.

`p'(x) = 3x^2 - 18x`

We now find the critical numbers, which occur when the derivative is `0` or is undefined. Since polynomials are continuous, we will only have critical numbers when `dy/dx = 0`.

`0 = 3x(x - 6)`

`x= 0 and 6`

We now test around the points to check whether the derivative is increasing or decreasing.

Test Point: `x = -1`

`p(-1) = 3(-1)^2 - 18(-1) = 3 + 18 = 21`

This means that `p(x)` is increasing on `(-oo, 0)`. Before confirming that `x= 0` is an absolute maximum though we must check the y-values of the function at the end points.

`p(-2) = (-2)^3 - 9(-2)^2 + 10`

`p(-2) = -8 -9(4) +10`

`p(-2) = -34`

This works!

`p(8) = 8^3 - 9(8)^2 + 10`

`p(8) = 512 - 576 + 10`

`p(8) = -54`

This works!

Therefore, `(0, 10)` is an absolute maximum on `[-2, 8]` (always specify, because on `(-oo, oo)`, this function has no absolute maximum).

Finally, we must ensure that `x = 6` is an absolute minimum.

Test Point `x = 5`

`p(5) = 5^3 - 9(5)^2 + 10`

`p(5) = -90`

We can now see that `p(x)` is decreasing on `(0, 6)`. This will mean that we have a minimum of some sort at `x = 6`. We know for sure that it will be either an absolute or local minimum because this critical point has a derivative equal to `0`, so the function changes vertical direction, or the derivative goes from decreasing to increasing.

`p(6)= 6^3 - 9(6)^2 + 10`

`p(6) = -98`

Since this is smaller than both `p(-2)` and `p(8)`, we can see that this is an absolute minimum.

Hopefully this helps!