SOLUTION: Find the absolute extrema of the function on the closed interval. f(x) = x^3 − 3/2x^2, [−3, 4] maximum (x, y) = Find the absolute extrema of the function o

Algebra ->  Trigonometry-basics -> SOLUTION: Find the absolute extrema of the function on the closed interval. f(x) = x^3 − 3/2x^2, [−3, 4] maximum (x, y) = Find the absolute extrema of the function o      Log On


   



Question 1187610: Find the absolute extrema of the function on the closed interval.
f(x) = x^3 − 3/2x^2, [−3, 4]
maximum (x, y) =

Find the absolute extrema of the function on the closed interval.
y = 3x^2/3 − 2x, [−1, 1]
maximum (x, y) =

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Find the absolute extrema of the function on the closed interval.
%22f%28x%29%22%22%22=%22%22x%5E3+%E2%88%92+expr%283%2F2%29x%5E2, [−3, 4]

The absolute extrema are either points at which the derivative is 0 or
undefined, or the endpoints of the interval, -3 or 4.

%22f%27%28x%29%22%22%22=%22%223x%5E2-3x which we set = 0

3x%5E2-3x%22%22=%22%220

3x%28x-3%29%22%22=%22%220



Points where derivative is 0 are (0,0) and (3,13.5)

Endpoints:



So the absolute maximum point is (4,40)
and the absolute minimum point is (-3,-40.5).
Find the absolute extrema of the function on the closed interval.
y = 3x^2/3 − 2x, [−1, 1]

Do it the same way all by yourself.  This time one of the absolute extrema
will be an an endpoint and the other at a point where the derivative is 0.

Edwin