# How do you find the absolute and local extreme values for #f(x)=(x-3)^2-9# on the interval [-8,-3]?

##### 1 Answer

Absolute Min:

Absolute Max:

#### Explanation:

We are interested in local and global extrema, so we will need to employ the first-derivative test.

Setting this equal to zero gives us the x-value(s) of our potential extrema, discounting our endpoints.

We will not check if

f(-8) = (-11)^2 - 9 = 121 - 9 = 112

f(-3) = (-6)^2 - 9 = 36 - 9 = 27

Thus, our function is decreasing throughout our interval. The function

graph{(x-3)^2 - 9 [-8, -3, 0, 120]}