Find the absolute maximum and minimum value of h(x)=2x^3+3x^2-72x+3 on [0,5]
+95 F'(X)=6x^2+6x-72
6x^2+6x-72=0
x^2+X-12=0
(X+4)(X-3)=0
x=-4 X=3
Then in a chart with X values on one side and f(X) values on the other side (I don't know how to make a chart on here if you can)
0, 3
5, -32
3, -132
So,
absolute maximum of f(X)=3 at X=0
absolute minimum of f(X)= -132 at X=3
