Question 1187610
Find the absolute extrema of the function on the closed interval.

{{{"f(x)"}}}{{{""=""}}}{{{x^3 − expr(3/2)x^2}}},    [−3, 4]<pre>

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

{{{"f'(x)"}}}{{{""=""}}}{{{3x^2-3x}}} which we set = 0

{{{3x^2-3x}}}{{{""=""}}}{{{0}}}

{{{3x(x-3)}}}{{{""=""}}}{{{0}}}

{{{matrix(5,3,

3x = 0,or,x-3=0,
x=0,"",x=3,
"f(0)"=0^3-expr(3/2)(0)^2,"","f(3)"=3^3-expr(3/2)(3)^2,
"f(0)"=0,"","f(3)"=27-27/2,
"","","f(3)"=27/2=13.5)}}}

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

Endpoints:

{{{matrix(5,3,

3x = 0,or,x-3=0,

x=-3,"",x=4,
"f(-3)"=(-3)^3-expr(3/2)(-3)^2, "", "f(4)"=4^3-expr(3/2)(4)^2,
"f(-3)"=-27-27/2,               "", "f(4)"=64-24,
"f(-3)"=-40.5,"","f(4)"=40)}}}

So the absolute maximum point is (4,40)
and the absolute minimum point is (-3,-40.5).
</pre>Find the absolute extrema of the function on the closed interval.<pre>
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</pre>