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

Apr 15, 2018

Absolute Min: $x = - 3$
Absolute Max: $x = - 8$

#### Explanation:

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

$f \left(x\right) = {\left(x - 3\right)}^{2} - 9 = {x}^{2} - 6 x + 9 - 9 = {x}^{2} - 6 x$
$f ' \left(x\right) = 2 x - 6$

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

$f ' \left(x\right) = 0 = 2 x - 6 \to 2 x = 6 \to x = 3$

We will not check if $x = 3$ is an extreme, since it is not in our interval. Because the first-derivative test did not yield any potential extremes in our interval, we know the function is either strictly decreasing or strictly increasing on this interval. As such, we need only to check our endpoints.

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 $f \left(x\right)$ has an absolute minimum as $x = - 3$ and an absolute maximum at $x = - 8$, on our interval. A graph of our function in this interval confirms.

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