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:
f'(5) = 6(5)^2 - 12(5) - 48 = 42
Test point 2:
f'(3) = 6(3)^2 - 12(3) - 48 = 54 - 36 - 48 = -30
So,
Test point 3:
f'(-1) = 6(-1)^2 - 12(-1) - 48 = 6 + 12 - 48 = -30
Test point 4:
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!