Question

How do you find the absolute maximum and minimum of the polynomial function of `f(x) = 2x^3 – 6x^2 – 48x + 24`?

Answer 1

We have:

•A local maximum at `(-2, 80)`
•A local minimum at `(4, -136)`

No absolute maximums or minimums.

Explanation:

First of all, by polynomial rules, there will be no absolute maximum or minimum. Since the highest degree term is of degree `3` (odd) and the coefficient is positive `(2)`, at left of the graph we will be at `(-x, -oo)` and work our way up as we go right towards `(x, oo)`. This means there will at most be a local max/min. To find these, we start by finding the derivative.

`f'(x) = 6x^2 - 12x - 48`

We must now find the critical numbers. These will contain our relative maximum and minimums. This is a polynomial function defined over all values of `x`. Our only critical point will come when the derivative equals `0`.

`0 = 6x^2 - 12x - 48`

`0 = 6(x^2 - 2x - 8)`

`0 = (x - 4)(x +2)`

`x = 4 and -2`

The next step is to check the sign of the derivative on both sides of the critical numbers. If `f'(x) < 0`, then `f(x)` is decreasing, but if `f'(x) > 0`, the `f(x)` will be increasing.

Test point 1: `x = 5`

`f'(5) = 6(5)^2 - 12(5) - 48 = 42`

Test point 2: `x = 3`

`f'(3) = 6(3)^2 - 12(3) - 48 = 54 - 36 - 48 = -30`

So, `x = 4` is a local minimum value.

Test point 3: `x = -1`

`f'(-1) = 6(-1)^2 - 12(-1) - 48 = 6 + 12 - 48 = -30`

Test point 4: `x = -3`

`f'(-3) = 6(-3)^2 - 12(-3) - 48 = 54 + 36 - 48 = 42`

Therefore, `x= -2` is a local maximum. Everything we have found algebraically agrees with what can be seen in the following graph of `f(x)`.

graph{2x^3- 6x^2 - 48x + 24 [-10, 10, -5, 5]}

Hopefully this helps!