Find the absolute maximum and absolute minimum values of the function
f(x)=(x-1)(x-5)^3+9
on each of the indicated intervals.
(A) Interval = [1, 4].
Absolute maximum =
Absolute minimum =
f(x)=(x-1)(x-5)^3+9
on each of the indicated intervals.
(A) Interval = [1, 4].
Absolute maximum =
Absolute minimum =
-
f(x) = (x-1)(x-5)^3 + 9
By the product rule, we have:
f '(x) = (x-5)^3 + 3(x-1)(x-5)^2.
Setting this equal to 0, gives:
(x-5)^2[(x-5) + 3(x-1)] = 0
x = 5 (let's disregard this value since it lies outside of our interval)
x - 5 + 3x - 3 = 0
4x - 8 = 0
4x = 8
x = 2.
We can disregard x = 5 since it doesn't lie within our interval [1,4]. All we've left to is to plug in our endpoints and points around our critical values into our derivative function to check the extrema, and then into our function to get our y-coordinates.
f '(1) = -216
f '(2) = Minimum (since the function is increasing)
f '(4) = 8
f (2) = -18
f (1) = 9
Aboslute Maximum = (1,9)
Absolute Minimum = (2,-18)
By the product rule, we have:
f '(x) = (x-5)^3 + 3(x-1)(x-5)^2.
Setting this equal to 0, gives:
(x-5)^2[(x-5) + 3(x-1)] = 0
x = 5 (let's disregard this value since it lies outside of our interval)
x - 5 + 3x - 3 = 0
4x - 8 = 0
4x = 8
x = 2.
We can disregard x = 5 since it doesn't lie within our interval [1,4]. All we've left to is to plug in our endpoints and points around our critical values into our derivative function to check the extrema, and then into our function to get our y-coordinates.
f '(1) = -216
f '(2) = Minimum (since the function is increasing)
f '(4) = 8
f (2) = -18
f (1) = 9
Aboslute Maximum = (1,9)
Absolute Minimum = (2,-18)