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

Question

Aurora Alvarado · Accepted Answer

You find the extremes by taking the derivative and setting it to #0# #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]}

Luke Alvarez · Answer

undefined To find the global extreme values of $ h(x) = x^2 - 3x $ on the interval $[0, 2]$, follow these steps:

1. Find the critical points of $ h(x) $ within the interval $[0, 2]$ by taking the derivative and setting it equal to zero:
$$ h'(x) = 2x - 3 $$
$$ 2x - 3 = 0 $$
$$ x = \frac{3}{2} $$

2. Evaluate $ h(x) $ at the critical point and at the endpoints of the interval:
$$ h(0) = (0)^2 - 3(0) = 0 $$
$$ h(2) = (2)^2 - 3(2) = -2 $$
$$ h\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) = \frac{9}{4} - \frac{9}{2} = -\frac{9}{4} $$

3. Compare the values obtained in step 2. The maximum value is the largest value, and the minimum value is the smallest value.

Therefore, the global maximum value of $ h(x) $ on the interval $[0, 2]$ is $ h(0) = 0 $, and the global minimum value is $ h\left(\frac{3}{2}\right) = -\frac{9}{4} $.