Question

Find the absolute maximum and absolute minimum values of f on the given interval. = f(x) = 2x3 – 3x2 – 72x + 3 [-4, 5] X (min) X (max)

Answer 1

The absolute maximum value is 211 and it occurs at x=-4, while the absolute minimum value is -205 and it occurs at x=4.

Step-by-step explanation:

To find the absolute maximum and minimum values of the function f(x) = 2x³ – 3x² – 72x + 3 on the interval [-4, 5], we can use the first derivative test and the endpoints of the interval to determine these values.

First, we need to find the critical points of the function by taking its derivative and setting it equal to zero. The derivative of f(x) is f’(x) = 6x² - 6x - 72. Setting this equal to zero gives us:

6x² - 6x - 72 = 0

We can factor out a 6 from the equation:

6(x² - x - 12) = 0

Then, we can factor the quadratic equation:

6(x - 4)(x + 3) = 0

This gives us two critical points: x = 4 and x = -3.

Next, we need to evaluate the function at these critical points and at the endpoints of the interval [-4, 5].

When x = -4:

f(-4) = 2(-4)³ – 3(-4)² – 72(-4) + 3

f(-4)= -128 + 48 + 288 + 3

f(-4) = 211

When x = 5:

f(5) = 2(5)³ – 3(5)² – 72(5) + 3

f(5) =250 -75 -360 +3

f(5) = -182

Now, we compare these values with the values at the critical points:

When x = -3:

f(-3) = 2(-3)³ – 3(-3)² – 72(-3) + 3

f(-3) = -54 -27 +216 +3

f(-3) =138

When x = 4:

f(4) = 2(4)³ – 3(4)² – 72(4) + 3

f(4) =128-48-288+3

f(4) =-205

From these calculations, we find that:

The absolute maximum value of f on the interval [-4,5] is 211 which occurs at x=-4. The absolute minimum value of f on the interval [-4,5] is -205 which occurs at x=4.

Answer 2

The absolute maximum is at x=−3 with a value of 136, and the absolute minimum is at x=4 with a value of -207.

To find the absolute maximum and minimum values of the function
`f(x)=2 x^3-3 x^2-72 x+1` on the interval [−4,5], we need to consider the critical points and endpoints of the interval.

Critical Points:

Find the critical points by setting the derivative of f(x) equal to zero and solving for x.

`f^(\prime)(x)=6 x^2-6 x-72`

Setting
`f^(\prime)(x)=0`, we get:

`6 x^2-6 x-72=0`

Factorizing, we get:

`(x+3)(x-4)=0`

So, x=−3 and x=4 are the critical points.

Endpoints:

Evaluate f(x) at the endpoints of the interval [−4,5], which are x=−4 and x=5.

Evaluate at Critical Points:

Evaluate f(x) at the critical points x=−3 and x=4.

Compare Values:

Compare the values of f(x) at the critical points and endpoints to determine the absolute maximum and minimum.

`\begin{array}{r}f(-4) \\f(-3) \\f(4) \\f(5)\end{array}`

The maximum value occurs at x=−3 with a value of 136, and the minimum value occurs at x=4 with a value of -207.