Math / Calculus

Question $f(x) = -2x^3 + 21x^2 - 36x$ on $[0, 7]$ a. Determine the absolute extreme values of $f$ on the given interval when they exist. b. Use a graphing utility to confirm your conclusions. Part 1 of 3

Studdy Solution

STEP 1

What is this asking? Find the highest and lowest points of a curvy line between two endpoints. Watch out! Don't forget to check the endpoints themselves, they can be sneaky and hold the extreme values!

STEP 2

1. Find the derivative.
2. Find critical points.
3. Evaluate at critical points and endpoints.

STEP 3

Alright, let's kick things off by finding the derivative of our function $f(x) = -2x^3 + 21x^2 - 36x$ .
Remember, the derivative tells us the slope of our function at any given point, which is key to finding those peaks and valleys!

STEP 4

Using the power rule, we bring down the exponent and multiply it by the coefficient, then decrease the exponent by one.
So, the derivative, denoted as $f'(x)$ , is: $f'(x) = -2 \cdot 3x^{3-1} + 21 \cdot 2x^{2-1} - 36 \cdot 1x^{1-1}$ $f'(x) = -6x^2 + 42x - 36$ Boom! There's our derivative!

STEP 5

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

STEP 6

Setting $f'(x)$ equal to zero gives us: $-6x^2 + 42x - 36 = 0$ We can simplify things by dividing everything by $-6$ : $x^2 - 7x + 6 = 0$ This is a classic quadratic equation, and we can factor it like a pro: $(x - 1)(x - 6) = 0$ This gives us two critical points: $x = 1$ and $x = 6$ .

STEP 7

Now for the grand finale!
We need to evaluate our original function $f(x)$ at our critical points ( $x = 1$ and $x = 6$ ) and the endpoints of our interval ( $x = 0$ and $x = 7$ ).

STEP 8

Let's start with $x = 0$ : $f(0) = -2(0)^3 + 21(0)^2 - 36(0) = 0$ Next, $x = 1$ : $f(1) = -2(1)^3 + 21(1)^2 - 36(1) = -2 + 21 - 36 = -17$ Now, $x = 6$ : $f(6) = -2(6)^3 + 21(6)^2 - 36(6) = -432 + 756 - 216 = 108$ Finally, $x = 7$ : $f(7) = -2(7)^3 + 21(7)^2 - 36(7) = -686 + 1029 - 252 = 91$

STEP 9

The absolute minimum value is $-17$ at $x = 1$ , and the absolute maximum value is $108$ at $x = 6$ .

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Question $f(x) = -2x^3 + 21x^2 - 36x$ on $[0, 7]$ a. Determine the absolute extreme values of $f$ on the given interval when they exist. b. Use a graphing utility to confirm your conclusions. Part 1 of 3

Studdy Solution

STEP 1

What is this asking? Find the highest and lowest points of a curvy line between two endpoints. Watch out! Don't forget to check the endpoints themselves, they can be sneaky and hold the extreme values!

STEP 2

1. Find the derivative.
2. Find critical points.
3. Evaluate at critical points and endpoints.

STEP 3

Alright, let's kick things off by finding the derivative of our function $f(x) = -2x^3 + 21x^2 - 36x$ .
Remember, the derivative tells us the slope of our function at any given point, which is key to finding those peaks and valleys!

STEP 4

Using the power rule, we bring down the exponent and multiply it by the coefficient, then decrease the exponent by one.
So, the derivative, denoted as $f'(x)$ , is: $f'(x) = -2 \cdot 3x^{3-1} + 21 \cdot 2x^{2-1} - 36 \cdot 1x^{1-1}$ $f'(x) = -6x^2 + 42x - 36$ Boom! There's our derivative!

STEP 5

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

STEP 6

Setting $f'(x)$ equal to zero gives us: $-6x^2 + 42x - 36 = 0$ We can simplify things by dividing everything by $-6$ : $x^2 - 7x + 6 = 0$ This is a classic quadratic equation, and we can factor it like a pro: $(x - 1)(x - 6) = 0$ This gives us two critical points: $x = 1$ and $x = 6$ .

STEP 7

Now for the grand finale!
We need to evaluate our original function $f(x)$ at our critical points ( $x = 1$ and $x = 6$ ) and the endpoints of our interval ( $x = 0$ and $x = 7$ ).

STEP 8

Let's start with $x = 0$ : $f(0) = -2(0)^3 + 21(0)^2 - 36(0) = 0$ Next, $x = 1$ : $f(1) = -2(1)^3 + 21(1)^2 - 36(1) = -2 + 21 - 36 = -17$ Now, $x = 6$ : $f(6) = -2(6)^3 + 21(6)^2 - 36(6) = -432 + 756 - 216 = 108$ Finally, $x = 7$ : $f(7) = -2(7)^3 + 21(7)^2 - 36(7) = -686 + 1029 - 252 = 91$

STEP 9

The absolute minimum value is $-17$ at $x = 1$ , and the absolute maximum value is $108$ at $x = 6$ .

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Questionf(x)=−2x3+21x2−36xf(x) = -2x^3 + 21x^2 - 36xf(x)=−2x3+21x2−36x on [0,7][0, 7][0,7] a. Determine the absolute extreme values of fff on the given interval when they exist. b. Use a graphing utility to confirm your conclusions. Part 1 of 3

Studdy Solution

STEP 1

What is this asking? Find the *highest* and *lowest* points of a curvy line between two endpoints. Watch out! Don't forget to check the *endpoints* themselves, they can be sneaky and hold the extreme values!

STEP 2

1. Find the derivative.2. Find critical points.3. Evaluate at critical points and endpoints.

STEP 3

Alright, let's **kick things off** by finding the derivative of our function f(x)=−2x3+21x2−36xf(x) = -2x^3 + 21x^2 - 36xf(x)=−2x3+21x2−36x.Remember, the derivative tells us the *slope* of our function at any given point, which is *key* to finding those peaks and valleys!

STEP 4

STEP 5

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

STEP 6

STEP 7

Now for the **grand finale**!We need to evaluate our original function f(x)f(x)f(x) at our critical points (x=1x = 1x=1 and x=6x = 6x=6) *and* the endpoints of our interval (x=0x = 0x=0 and x=7x = 7x=7).

STEP 8

STEP 9

The **absolute minimum** value is −17-17−17 at x=1x = 1x=1, and the **absolute maximum** value is 108108108 at x=6x = 6x=6.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question $f(x) = -2x^3 + 21x^2 - 36x$ on $[0, 7]$ a. Determine the absolute extreme values of $f$ on the given interval when they exist. b. Use a graphing utility to confirm your conclusions. Part 1 of 3

What is this asking? Find the highest and lowest points of a curvy line between two endpoints. Watch out! Don't forget to check the endpoints themselves, they can be sneaky and hold the extreme values!

1. Find the derivative.
2. Find critical points.
3. Evaluate at critical points and endpoints.

Alright, let's kick things off by finding the derivative of our function $f(x) = -2x^3 + 21x^2 - 36x$ .
Remember, the derivative tells us the slope of our function at any given point, which is key to finding those peaks and valleys!

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

Now for the grand finale!
We need to evaluate our original function $f(x)$ at our critical points ( $x = 1$ and $x = 6$ ) and the endpoints of our interval ( $x = 0$ and $x = 7$ ).

The absolute minimum value is $-17$ at $x = 1$ , and the absolute maximum value is $108$ at $x = 6$ .