Find the absolute maximum and absolute minimum values of the function

f(x)=(x−2)(x−5)^3+11 on each of the indicated intervals.
Enter -1000 for any absolute extrema that does not exist.

(A) Interval = [1,4]
Absolute maximum =

Absolute minimum =

(B) Interval = [1,8] .
Absolute maximum =

Absolute minimum =

(C) Interval = [4,9] .
Absolute maximum =

Absolute minimum =

My answers are:
(A) Interval = [1,4]
Absolute maximum = 9

Absolute minimum = -5

(B) Interval = [1,8] .
Absolute maximum = 173

Absolute minimum = -5

(C) Interval = [4,9] .
Absolute maximum = 459

Absolute minimum = -1000

Only the max for the second one and the max for the third one are right.... I don't know what to do....

f'(x)=(x-5)^3+(x-2)3(x-5)^2=(4x-11)(x-5)^2

(A) Absmax=f(1)=64+11=75
Absmin=f(11/4)=-3^7/2^8+11=-2187/256+
+11=629/256
(B) Absmax=f(8)=173
Absmin=f(11/4)=629/256
(C) Absmax=f(9)=459
Absmin=f(4)=9

To find the absolute maximum and minimum values of the given function on each interval, we need to check the critical points and endpoints of the interval.

Let's go step by step:

(A) Interval = [1,4]

1. Find critical points by setting the derivative of the function equal to zero and solving for x:
f'(x) = 4(x-5)^3 + (x-2)(3(x-5)^2)
Setting f'(x) = 0 and solving:
4(x-5)^3 + 3(x-2)(x-5)^2 = 0
(x-5)^2[4(x-5) + 3(x-2)] = 0
(x-5)[4(x-5) + 3(x-2)] = 0
(x-5)(4x - 20 + 3x - 6) = 0
(x-5)(7x - 26) = 0
Solving each factor separately:
x-5 = 0 -> x = 5
7x - 26 = 0 -> x = 26/7

2. Check the function value at the critical points and endpoints:
f(1) = (1-2)(1-5)^3 + 11 = (-1)(-4)^3 + 11 = -59
f(4) = (4-2)(4-5)^3 + 11 = (2)(-1)^3 + 11 = 9
f(5) = (5-2)(5-5)^3 + 11 = (3)(0)^3 + 11 = 11

Based on the calculations, the absolute maximum on the interval [1,4] is 11 (at x = 5) and the absolute minimum is -59 (at x = 1).

(B) Interval = [1,8]

1. Find critical points by setting the derivative of the function equal to zero and solving for x:
Using the derivative we found in step (A), f'(x) = (x-5)(7x-26)

2. Check the function value at the critical points and endpoints:
f(1) = -59
f(8) = (8-2)(8-5)^3 + 11 = (6)(3)^3 + 11 = 173
f(5) = 11
Based on the calculations, the absolute maximum on the interval [1,8] is 173 (at x = 8) and the absolute minimum is -59 (at x = 1).

(C) Interval = [4,9]

1. Find critical points by setting the derivative of the function equal to zero and solving for x:
Using the derivative we found in step (A), f'(x) = (x-5)(7x-26)

2. Check the function value at the critical points and endpoints:
f(4) = 9
f(9) = (9-2)(9-5)^3 + 11 = (7)(4)^3 + 11 = 459
f(5) = 11
Based on the calculations, the absolute maximum on the interval [4,9] is 459 (at x = 9) and the absolute minimum does not exist (we used -1000 instead).

Therefore, the corrected answers are:

(A) Interval = [1,4]
Absolute maximum = 11
Absolute minimum = -59

(B) Interval = [1,8]
Absolute maximum = 173
Absolute minimum = -59

(C) Interval = [4,9]
Absolute maximum = 459
Absolute minimum = -1000 (does not exist)

To find the absolute maximum and absolute minimum values of a function on a given interval, we can use the process of finding critical points and evaluating the function at these points, as well as the endpoints of the interval.

Let's go through each interval one by one and find the absolute maximum and absolute minimum values:

(A) Interval = [1,4]

1. Find the critical points:
To find the critical points, we need to find where the derivative of the function is equal to zero or does not exist. So, let's find the derivative of the function, f'(x):

f(x) = (x - 2)(x - 5)^3 + 11
f'(x) = (x - 5)^3 + 3(x - 2)(x - 5)^2

To find where f'(x) = 0, we solve the equation:
(x - 5)^3 + 3(x - 2)(x - 5)^2 = 0

Simplifying the equation, we get:
(x - 5)^2 [(x - 5) + 3(x - 2)] = 0
(x - 5)^2 (x - 2x + 10) = 0
(x - 5)^2 (x + 8) = 0

(x - 5) = 0 or (x + 8) = 0
x = 5 or x = -8

So, the critical points are x = 5 and x = -8.

2. Evaluate the function at the critical points and the endpoints of the interval:
f(1) = (1 - 2)(1 - 5)^3 + 11 = -197
f(4) = (4 - 2)(4 - 5)^3 + 11 = -77
f(5) = (5 - 2)(5 - 5)^3 + 11 = 11
f(-8) = (-8 - 2)(-8 - 5)^3 + 11 = -1175

3. Compare the values obtained:
The absolute maximum value is the highest value obtained, which is 11.
The absolute minimum value is the lowest value obtained, which is -1175.

So, the correct answers for (A) Interval = [1,4] are:
Absolute maximum = 11
Absolute minimum = -1175

Now let's move on to the next interval.

(B) Interval = [1,8]

1. Find the critical points:
We already found the critical points in the previous step, which are x = 5 and x = -8.

2. Evaluate the function at the critical points and the endpoints of the interval:
f(1) = -197
f(4) = -77
f(5) = 11
f(8) = (8 - 2)(8 - 5)^3 + 11 = 173

3. Compare the values obtained:
The absolute maximum value is 173.
The absolute minimum value is -197.

So, the correct answers for (B) Interval = [1,8] are:
Absolute maximum = 173
Absolute minimum = -197

Finally, let's move on to the last interval.

(C) Interval = [4,9]

1. Find the critical points:
Again, we have x = 5 and x = -8 as the critical points.

2. Evaluate the function at the critical points and the endpoints of the interval:
f(4) = -77
f(5) = 11
f(8) = 173
f(9) = (9 - 2)(9 - 5)^3 + 11 = 459

3. Compare the values obtained:
The absolute maximum value is 459.
The absolute minimum value is -77.

So, the correct answers for (C) Interval = [4,9] are:
Absolute maximum = 459
Absolute minimum = -77

Therefore, the correct answers are:
(A) Interval = [1,4]
Absolute maximum = 11
Absolute minimum = -1175

(B) Interval = [1,8]
Absolute maximum = 173
Absolute minimum = -197

(C) Interval = [4,9]
Absolute maximum = 459
Absolute minimum = -77