Question

Find the values of x where the absolute extrema (that is, absolute maximum and
absolute minimum) if they exist.
(b) Find the absolute extrema if they exist.
1. f(x) = x^3 −6x^2 + 9x −8, 0 ≤ x ≤ 5
2. f(x) = x^3 −3x^2 −24x + 5, −3 ≤ x ≤ 6
3. f(x) = 1/3x^3 + 3/2x^2 −4x + 1, −5 ≤ x ≤ 2
4. f(x) = 12 −x − 9/x, 1 ≤ x ≤ 5
Answers: Provided below, I need a detailed explanation on how to get to these answers.
SHOW WORK
1. Absolute maximum of 12 at x=5; absolute minimum of -8 at x=0 and x=3.
2. Absolute maximum of 33 at x=-2; absolute minimum of -75 at x=4
3. Absolute maximum of 19.67 at x=-4; absolute minimum of -1.17 at x= 1
4. Critical value on the interval [1,5] is x=3. Notice that though x=-3 is a critical value, it is not part of the interval [1,5] and is therefore ignored. The absolute maximum is 6 and occurs at x=3; The absolute minimum is 2 and occurs at
x=1

Answer 1

The value of the absolute extrema of the functions are

1. f(x) = x³ −6x² + 9x −8, maximum is 12 and the minimum is -8

2. f(x) = x³ −3x² −24x + 5 maximum is 101 and the minimum is -43

3. f(x) = 1/3x³ + 3/2x² −4x + 1 maximum is 1.67 and the minimum is 16.83

4. f(x) = 12 −x − 9/x maximum is 5.2 and the minimum is 2

The term absolute extrema means the point at which a maximum or minimum value of the function is obtained.

Here we have given the following functions

1. f(x) = x³ −6x² + 9x −8, 0 ≤ x ≤ 5

2. f(x) =  x³ −3x² −24x + 5 , −3 ≤ x ≤ 6

3. f(x) = 1/3x³ + 3/2x² −4x + 1 , −5 ≤ x ≤ 2

4. f(x) = 12 −x − 9/x, 1 ≤ x ≤ 5

And we need to find the absolute extrema if they exist.

In order to find the absolute maximum and minimum we have to take the maximum and minimum limit of each function. And then we have to apply the values on the function to find the exact value of it.

For the first function the maximum limit 5 and the minimum limit is 0.

Therefore, the maximum value of the function is calculated as,

=> f(5) = (5)³ −6(5)² + 9(5) −8

=> f(5) = 125 - 6(25) + 45 - 8

=> f(5) = 125 - 150 + 45 - 8

=> f(5) = 12

And the minimum value of the function is calculated as,

=> f(0) = (0)³ −6(0)² + 9(0) −8

=> f(0) = (0) - 6(0) + 0 - 8

=> f(0) = 0 - 0 + 0 - 8

=> f(0) = -8

Similarly, the next function has the maximum limit 6 and the minimum limit as -3,

Therefore, the maximum value of the function is calculated as,

=> f(6) = (6)³ −3(6)² −2(6) + 5

=> f(6) = 216 - 3(36) - 12 + 5

=> f(6) = 216 - 108 - 12 + 5

=> f(6) = 101

And the minimum value of the function is calculated as,

=> f(-3) = (-3)³ −3(-3)² −2(-3) + 5

=> f(-3) = -27 - 27 + 6 + 5

=> f(-3) = -43

Then the next function has the maximum limit 2 and the minimum limit as -5,

Therefore, the maximum value of the function is calculated as,

=> f(2) = 1/3(2)³ + 3/2(2)² −4(2) + 1

=> f(2) = 8/3 + 6 - 8 + 1

=> f(2) = 1.67

And the minimum value of the function is calculated as,

=> f(-5) = 1/3(-5)³ + 3/2(-5)² −4(-5) + 1

=> f(-5) = -125/3 + 75/2 + 20 + 1

=> f(-5) = 16.83

Finally, the function has the maximum limit 5 and the minimum limit as 1,

Therefore, the maximum value of the function is calculated as,

=> f(5) = 12 − 5 − 9/5

=> f(5) = 5.2

And the minimum value of the function is calculated as,

=> f(1) = 12 − 1 − 9/1

=> f(1) = 2

To know more about absolute extrema here.

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