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!