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

1 Answer
Feb 21, 2017

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!