Math / Calculus

QuestionConsider the function $f(x)=2 x^{3}+21 x^{2}-48 x+11, \quad-8 \leq x \leq 2$ . Use the derivatives to algebraically answer the question:
This function has an absolutelminimum value equal to $\square$ and an absolute maximum value equal to $\square$ Question Help: Message instructor Submit Question Jump to Answer

Studdy Solution

STEP 1

What is this asking? Find the highest and lowest points of a curvy line between $-8$ and $2$ . Watch out! Don't forget to check the endpoints, they can be sneaky and hide the highest or lowest points!

STEP 2

1. Find the derivative.
2. Find critical points.
3. Check endpoints and critical points.

STEP 3

Let's kick things off by finding the derivative of our function $f(x) = 2x^3 + 21x^2 - 48x + 11$ .
Remember, the derivative tells us the instantaneous rate of change, like the slope at any given point.
The power rule is our weapon of choice here!

STEP 4

Bring down the power and multiply it with the coefficient, then reduce the power by one.
Doing this for each term, we get: $f'(x) = 3 \cdot 2x^{3-1} + 2 \cdot 21x^{2-1} - 1 \cdot 48x^{1-1} + 0 = 6x^2 + 42x - 48.$ So, our shiny new derivative is $f'(x) = 6x^2 + 42x - 48$ .

STEP 5

Now, let's hunt down those critical points!
Critical points happen where the derivative is zero or undefined.
Our derivative is a nice, well-behaved polynomial, so it's defined everywhere.
We just need to find where it's equal to zero.

STEP 6

Setting $f'(x) = 0$ , we get $6x^2 + 42x - 48 = 0$ .
We can wrangle this equation a bit by dividing everything by $6$ , giving us $x^2 + 7x - 8 = 0$ .

STEP 7

This quadratic equation can be factored as $(x+8)(x-1) = 0$ .
This gives us two critical points: $x = -8$ and $x = 1$ .

STEP 8

Alright, we've got our critical points $x = -8$ and $x = 1$ , and our endpoints are $x = -8$ and $x = 2$ .
Let's plug these values back into the original function $f(x)$ to see what we get!

STEP 9

For $x = -8$ , we have $f(-8) = 2(-8)^3 + 21(-8)^2 - 48(-8) + 11 = -1024 + 1344 + 384 + 11 = \textbf{715}$ .

STEP 10

For $x = 1$ , we have $f(1) = 2(1)^3 + 21(1)^2 - 48(1) + 11 = 2 + 21 - 48 + 11 = \textbf{-14}$ .

STEP 11

For $x = 2$ , we have $f(2) = 2(2)^3 + 21(2)^2 - 48(2) + 11 = 16 + 84 - 96 + 11 = \textbf{15}$ .

STEP 12

The absolute minimum value is $-14$ at $x=1$ , and the absolute maximum value is $715$ at $x=-8$ .

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Studdy Solution

STEP 1

What is this asking? Find the highest and lowest points of a curvy line between $-8$ and $2$ . Watch out! Don't forget to check the endpoints, they can be sneaky and hide the highest or lowest points!

STEP 2

1. Find the derivative.
2. Find critical points.
3. Check endpoints and critical points.

STEP 3

Let's kick things off by finding the derivative of our function $f(x) = 2x^3 + 21x^2 - 48x + 11$ .
Remember, the derivative tells us the instantaneous rate of change, like the slope at any given point.
The power rule is our weapon of choice here!

STEP 4

Bring down the power and multiply it with the coefficient, then reduce the power by one.
Doing this for each term, we get: $f'(x) = 3 \cdot 2x^{3-1} + 2 \cdot 21x^{2-1} - 1 \cdot 48x^{1-1} + 0 = 6x^2 + 42x - 48.$ So, our shiny new derivative is $f'(x) = 6x^2 + 42x - 48$ .

STEP 5

Now, let's hunt down those critical points!
Critical points happen where the derivative is zero or undefined.
Our derivative is a nice, well-behaved polynomial, so it's defined everywhere.
We just need to find where it's equal to zero.

STEP 6

Setting $f'(x) = 0$ , we get $6x^2 + 42x - 48 = 0$ .
We can wrangle this equation a bit by dividing everything by $6$ , giving us $x^2 + 7x - 8 = 0$ .

STEP 7

This quadratic equation can be factored as $(x+8)(x-1) = 0$ .
This gives us two critical points: $x = -8$ and $x = 1$ .

STEP 8

Alright, we've got our critical points $x = -8$ and $x = 1$ , and our endpoints are $x = -8$ and $x = 2$ .
Let's plug these values back into the original function $f(x)$ to see what we get!

STEP 9

For $x = -8$ , we have $f(-8) = 2(-8)^3 + 21(-8)^2 - 48(-8) + 11 = -1024 + 1344 + 384 + 11 = \textbf{715}$ .

STEP 10

For $x = 1$ , we have $f(1) = 2(1)^3 + 21(1)^2 - 48(1) + 11 = 2 + 21 - 48 + 11 = \textbf{-14}$ .

STEP 11

For $x = 2$ , we have $f(2) = 2(2)^3 + 21(2)^2 - 48(2) + 11 = 16 + 84 - 96 + 11 = \textbf{15}$ .

STEP 12

The absolute minimum value is $-14$ at $x=1$ , and the absolute maximum value is $715$ at $x=-8$ .

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Studdy Solution

STEP 1

What is this asking? Find the highest and lowest points of a curvy line between −8-8−8 and 222. Watch out! Don't forget to check the endpoints, they can be sneaky and hide the highest or lowest points!

STEP 2

1. Find the derivative.2. Find critical points.3. Check endpoints and critical points.

STEP 3

Let's **kick things off** by finding the derivative of our function f(x)=2x3+21x2−48x+11f(x) = 2x^3 + 21x^2 - 48x + 11f(x)=2x3+21x2−48x+11.Remember, the derivative tells us the *instantaneous rate of change*, like the slope at any given point.The **power rule** is our weapon of choice here!

STEP 4

STEP 5

Now, let's **hunt down** those critical points!Critical points happen where the derivative is zero or undefined.Our derivative is a nice, well-behaved polynomial, so it's defined everywhere.We just need to find where it's equal to zero.

STEP 6

Setting f′(x)=0f'(x) = 0f′(x)=0, we get 6x2+42x−48=06x^2 + 42x - 48 = 06x2+42x−48=0.We can **wrangle this equation** a bit by dividing everything by 666, giving us x2+7x−8=0x^2 + 7x - 8 = 0x2+7x−8=0.

STEP 7

This quadratic equation can be factored as (x+8)(x−1)=0(x+8)(x-1) = 0(x+8)(x−1)=0.This gives us two **critical points**: x=−8x = -8x=−8 and x=1x = 1x=1.

STEP 8

Alright, we've got our **critical points** x=−8x = -8x=−8 and x=1x = 1x=1, and our **endpoints** are x=−8x = -8x=−8 and x=2x = 2x=2.Let's plug these values back into the *original function* f(x)f(x)f(x) to see what we get!

STEP 9

For x=−8x = -8x=−8, we have f(−8)=2(−8)3+21(−8)2−48(−8)+11=−1024+1344+384+11=715f(-8) = 2(-8)^3 + 21(-8)^2 - 48(-8) + 11 = -1024 + 1344 + 384 + 11 = \textbf{715}f(−8)=2(−8)3+21(−8)2−48(−8)+11=−1024+1344+384+11=715.

STEP 10

For x=1x = 1x=1, we have f(1)=2(1)3+21(1)2−48(1)+11=2+21−48+11=-14f(1) = 2(1)^3 + 21(1)^2 - 48(1) + 11 = 2 + 21 - 48 + 11 = \textbf{-14}f(1)=2(1)3+21(1)2−48(1)+11=2+21−48+11=-14.

STEP 11

For x=2x = 2x=2, we have f(2)=2(2)3+21(2)2−48(2)+11=16+84−96+11=15f(2) = 2(2)^3 + 21(2)^2 - 48(2) + 11 = 16 + 84 - 96 + 11 = \textbf{15}f(2)=2(2)3+21(2)2−48(2)+11=16+84−96+11=15.

STEP 12

The **absolute minimum** value is −14-14−14 at x=1x=1x=1, and the **absolute maximum** value is 715715715 at x=−8x=-8x=−8.

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What is this asking? Find the highest and lowest points of a curvy line between $-8$ and $2$ . Watch out! Don't forget to check the endpoints, they can be sneaky and hide the highest or lowest points!

1. Find the derivative.
2. Find critical points.
3. Check endpoints and critical points.

Let's kick things off by finding the derivative of our function $f(x) = 2x^3 + 21x^2 - 48x + 11$ .
Remember, the derivative tells us the instantaneous rate of change, like the slope at any given point.
The power rule is our weapon of choice here!

Now, let's hunt down those critical points!
Critical points happen where the derivative is zero or undefined.
Our derivative is a nice, well-behaved polynomial, so it's defined everywhere.
We just need to find where it's equal to zero.

Setting $f'(x) = 0$ , we get $6x^2 + 42x - 48 = 0$ .
We can wrangle this equation a bit by dividing everything by $6$ , giving us $x^2 + 7x - 8 = 0$ .

This quadratic equation can be factored as $(x+8)(x-1) = 0$ .
This gives us two critical points: $x = -8$ and $x = 1$ .

Alright, we've got our critical points $x = -8$ and $x = 1$ , and our endpoints are $x = -8$ and $x = 2$ .
Let's plug these values back into the original function $f(x)$ to see what we get!

For $x = -8$ , we have $f(-8) = 2(-8)^3 + 21(-8)^2 - 48(-8) + 11 = -1024 + 1344 + 384 + 11 = \textbf{715}$ .

For $x = 1$ , we have $f(1) = 2(1)^3 + 21(1)^2 - 48(1) + 11 = 2 + 21 - 48 + 11 = \textbf{-14}$ .

For $x = 2$ , we have $f(2) = 2(2)^3 + 21(2)^2 - 48(2) + 11 = 16 + 84 - 96 + 11 = \textbf{15}$ .

The absolute minimum value is $-14$ at $x=1$ , and the absolute maximum value is $715$ at $x=-8$ .