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]}