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

Jul 27, 2015

You find the extremes by taking the derivative and setting it to $0$
$h ' \left(x\right) = 2 x - 3 = 0 \to x = 1 \frac{1}{2} \to h \left(x\right) = - 2 \frac{1}{4} \to \left(1 \frac{1}{2} , - 2 \frac{1}{4}\right)$
At $x = 0 \to h \left(x\right) = 0 \to \left(0 , 0\right)$
will be the maximum in the interval $\left[0 , 2\right]$, but there is no maximumfor the function as a whole.