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

Question

Luke Alvarez · Accepted Answer

Absolute Min: #x = -3#
Absolute Max: #x = -8# We are interested in local and global extrema, so we will need to employ the first-derivative test. #f(x) = (x-3)^2 - 9 = x^2 - 6x + 9 - 9 = x^2 - 6x# #f'(x) = 2x - 6# Setting this equal to zero gives us the x-value(s) of our potential extrema, discounting our endpoints. #f'(x) = 0 = 2x - 6 -> 2x = 6 -> 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(x)# 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]}

Cameron Alexander · Answer

undefined To find the absolute and local extreme values of the function $f(x) = (x - 3)^2 - 9$ on the interval $[-8, -3]$, we first find the critical points by setting the derivative of $f(x)$ equal to zero and solving for $x$. Then, we evaluate $f(x)$ at the critical points and at the endpoints of the interval to determine the extreme values.

1. Find the derivative of $f(x)$:
$$f'(x) = 2(x - 3)$$

2. Set $f'(x)$ equal to zero and solve for $x$ to find the critical points:
$$2(x - 3) = 0$$
$$x - 3 = 0$$
$$x = 3$$

3. Evaluate $f(x)$ at the critical point and at the endpoints of the interval:
- $f(-8) = (-8 - 3)^2 - 9 = (-11)^2 - 9 = 121 - 9 = 112$
- $f(-3) = (-3 - 3)^2 - 9 = (-6)^2 - 9 = 36 - 9 = 27$
- $f(3) = (3 - 3)^2 - 9 = 0^2 - 9 = -9$

4. Compare the values of $f(x)$ at the critical point and at the endpoints to find the absolute and local extreme values:
- Absolute minimum value: $f(-8) = 112$
- Absolute maximum value: $f(-3) = 27$
- Local minimum value: $f(3) = -9$