Question

How do you find the global extreme values for `h(x)= x^2-3x` on [0,2]?

Answer 1

You find the extremes by taking the derivative and setting it to `0`

Explanation:

`h'(x)=2x-3=0->x=1 1/2->h(x)=-2 1/4->(1 1/2,-2 1/4)`
which is a minimum

At `x=0->h(x)=0->(0,0)`
will be the maximum in the interval `[0,2]`, but there is no maximumfor the function as a whole.
graph{x^2-3x [-7.9, 7.9, -3.95, 3.95]}