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

##### 1 Answer

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

#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

#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

**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,

**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,

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

Hopefully this helps!