Math / Calculus

QuestionScore Part VI Solve the following questions. (4 scores per question. The total is 8 scores.)
19. Consider $y=\frac{1}{3} x^{3}-x+1$ on the interval $[0,3]$ , find the absolute maximum and minimum values of y .

Studdy Solution

STEP 1

Assumptions
1. The function given is $y = \frac{1}{3}x^3 - x + 1$ .
2. We need to find the absolute maximum and minimum values of $y$ on the interval $[0, 3]$ .
3. The function is continuous on the closed interval $[0, 3]$ .

STEP 2

To find the absolute maximum and minimum values of a continuous function on a closed interval, we need to evaluate the function at critical points and at the endpoints of the interval.

STEP 3

First, find the derivative of the function $y = \frac{1}{3}x^3 - x + 1$ .
$y' = \frac{d}{dx} \left(\frac{1}{3}x^3 - x + 1\right)$

STEP 4

Calculate the derivative.
$y' = x^2 - 1$

STEP 5

Find the critical points by setting the derivative equal to zero and solving for $x$ .
$x^2 - 1 = 0$

STEP 6

Solve the equation $x^2 - 1 = 0$ .
$x^2 = 1$

STEP 7

Take the square root of both sides to find the critical points.
$x = \pm 1$

STEP 8

Since we are only interested in the interval $[0, 3]$ , consider only the critical point $x = 1$ (as $x = -1$ is outside the interval).

STEP 9

Evaluate the function $y$ at the critical point $x = 1$ .
$y(1) = \frac{1}{3}(1)^3 - 1 + 1$

STEP 10

Calculate $y(1)$ .
$y(1) = \frac{1}{3} - 1 + 1 = \frac{1}{3}$

STEP 11

Evaluate the function $y$ at the endpoints of the interval, $x = 0$ and $x = 3$ .
$y(0) = \frac{1}{3}(0)^3 - 0 + 1$

STEP 12

Calculate $y(0)$ .
$y(0) = 1$

STEP 13

Evaluate the function at the other endpoint $x = 3$ .
$y(3) = \frac{1}{3}(3)^3 - 3 + 1$

STEP 14

Calculate $y(3)$ .
$y(3) = \frac{1}{3}(27) - 3 + 1 = 9 - 3 + 1 = 7$

STEP 15

Compare the values of $y$ at the critical point and the endpoints: $y(0) = 1$ , $y(1) = \frac{1}{3}$ , $y(3) = 7$ .

STEP 16

Determine the absolute maximum and minimum values.
- The absolute maximum value is $7$ at $x = 3$ . - The absolute minimum value is $\frac{1}{3}$ at $x = 1$ .

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QuestionScore Part VI Solve the following questions. (4 scores per question. The total is 8 scores.)
19. Consider $y=\frac{1}{3} x^{3}-x+1$ on the interval $[0,3]$ , find the absolute maximum and minimum values of y .

Studdy Solution

STEP 1

Assumptions
1. The function given is $y = \frac{1}{3}x^3 - x + 1$ .
2. We need to find the absolute maximum and minimum values of $y$ on the interval $[0, 3]$ .
3. The function is continuous on the closed interval $[0, 3]$ .

STEP 2

To find the absolute maximum and minimum values of a continuous function on a closed interval, we need to evaluate the function at critical points and at the endpoints of the interval.

STEP 3

First, find the derivative of the function $y = \frac{1}{3}x^3 - x + 1$ .
$y' = \frac{d}{dx} \left(\frac{1}{3}x^3 - x + 1\right)$

STEP 4

Calculate the derivative.
$y' = x^2 - 1$

STEP 5

Find the critical points by setting the derivative equal to zero and solving for $x$ .
$x^2 - 1 = 0$

STEP 6

Solve the equation $x^2 - 1 = 0$ .
$x^2 = 1$

STEP 7

Take the square root of both sides to find the critical points.
$x = \pm 1$

STEP 8

Since we are only interested in the interval $[0, 3]$ , consider only the critical point $x = 1$ (as $x = -1$ is outside the interval).

STEP 9

Evaluate the function $y$ at the critical point $x = 1$ .
$y(1) = \frac{1}{3}(1)^3 - 1 + 1$

STEP 10

Calculate $y(1)$ .
$y(1) = \frac{1}{3} - 1 + 1 = \frac{1}{3}$

STEP 11

Evaluate the function $y$ at the endpoints of the interval, $x = 0$ and $x = 3$ .
$y(0) = \frac{1}{3}(0)^3 - 0 + 1$

STEP 12

Calculate $y(0)$ .
$y(0) = 1$

STEP 13

Evaluate the function at the other endpoint $x = 3$ .
$y(3) = \frac{1}{3}(3)^3 - 3 + 1$

STEP 14

Calculate $y(3)$ .
$y(3) = \frac{1}{3}(27) - 3 + 1 = 9 - 3 + 1 = 7$

STEP 15

Compare the values of $y$ at the critical point and the endpoints: $y(0) = 1$ , $y(1) = \frac{1}{3}$ , $y(3) = 7$ .

STEP 16

Determine the absolute maximum and minimum values.
- The absolute maximum value is $7$ at $x = 3$ . - The absolute minimum value is $\frac{1}{3}$ at $x = 1$ .

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QuestionScore Part VI Solve the following questions. (4 scores per question. The total is 8 scores.)19. Consider y=13x3−x+1y=\frac{1}{3} x^{3}-x+1y=31​x3−x+1 on the interval [0,3][0,3][0,3], find the absolute maximum and minimum values of y .

Studdy Solution

STEP 1

Assumptions1. The function given is y=13x3−x+1 y = \frac{1}{3}x^3 - x + 1 y=31​x3−x+1.2. We need to find the absolute maximum and minimum values of y y y on the interval [0,3][0, 3][0,3].3. The function is continuous on the closed interval [0,3][0, 3][0,3].

STEP 2

To find the absolute maximum and minimum values of a continuous function on a closed interval, we need to evaluate the function at critical points and at the endpoints of the interval.

STEP 3

First, find the derivative of the function y=13x3−x+1 y = \frac{1}{3}x^3 - x + 1 y=31​x3−x+1.y′=ddx(13x3−x+1) y' = \frac{d}{dx} \left(\frac{1}{3}x^3 - x + 1\right) y′=dxd​(31​x3−x+1)

STEP 4

Calculate the derivative.y′=x2−1 y' = x^2 - 1 y′=x2−1

STEP 5

Find the critical points by setting the derivative equal to zero and solving for x x x.x2−1=0 x^2 - 1 = 0 x2−1=0

STEP 6

Solve the equation x2−1=0 x^2 - 1 = 0 x2−1=0.x2=1 x^2 = 1 x2=1

STEP 7

Take the square root of both sides to find the critical points.x=±1 x = \pm 1 x=±1

STEP 8

Since we are only interested in the interval [0,3][0, 3][0,3], consider only the critical point x=1 x = 1 x=1 (as x=−1 x = -1 x=−1 is outside the interval).

STEP 9

Evaluate the function y y y at the critical point x=1 x = 1 x=1.y(1)=13(1)3−1+1 y(1) = \frac{1}{3}(1)^3 - 1 + 1 y(1)=31​(1)3−1+1

STEP 10

Calculate y(1) y(1) y(1).y(1)=13−1+1=13 y(1) = \frac{1}{3} - 1 + 1 = \frac{1}{3} y(1)=31​−1+1=31​

STEP 11

Evaluate the function y y y at the endpoints of the interval, x=0 x = 0 x=0 and x=3 x = 3 x=3.y(0)=13(0)3−0+1 y(0) = \frac{1}{3}(0)^3 - 0 + 1 y(0)=31​(0)3−0+1

STEP 12

Calculate y(0) y(0) y(0).y(0)=1 y(0) = 1 y(0)=1

STEP 13

Evaluate the function at the other endpoint x=3 x = 3 x=3.y(3)=13(3)3−3+1 y(3) = \frac{1}{3}(3)^3 - 3 + 1 y(3)=31​(3)3−3+1

STEP 14

Calculate y(3) y(3) y(3).y(3)=13(27)−3+1=9−3+1=7 y(3) = \frac{1}{3}(27) - 3 + 1 = 9 - 3 + 1 = 7 y(3)=31​(27)−3+1=9−3+1=7

STEP 15

Compare the values of y y y at the critical point and the endpoints: y(0)=1 y(0) = 1 y(0)=1, y(1)=13 y(1) = \frac{1}{3} y(1)=31​, y(3)=7 y(3) = 7 y(3)=7.

STEP 16

Determine the absolute maximum and minimum values.- The absolute maximum value is 7 7 7 at x=3 x = 3 x=3. - The absolute minimum value is 13 \frac{1}{3} 31​ at x=1 x = 1 x=1.

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QuestionScore Part VI Solve the following questions. (4 scores per question. The total is 8 scores.)
19. Consider $y=\frac{1}{3} x^{3}-x+1$ on the interval $[0,3]$ , find the absolute maximum and minimum values of y .

Assumptions
1. The function given is $y = \frac{1}{3}x^3 - x + 1$ .
2. We need to find the absolute maximum and minimum values of $y$ on the interval $[0, 3]$ .
3. The function is continuous on the closed interval $[0, 3]$ .

First, find the derivative of the function $y = \frac{1}{3}x^3 - x + 1$ .
$y' = \frac{d}{dx} \left(\frac{1}{3}x^3 - x + 1\right)$

Calculate the derivative.
$y' = x^2 - 1$

Find the critical points by setting the derivative equal to zero and solving for $x$ .
$x^2 - 1 = 0$

Solve the equation $x^2 - 1 = 0$ .
$x^2 = 1$

Take the square root of both sides to find the critical points.
$x = \pm 1$

Since we are only interested in the interval $[0, 3]$ , consider only the critical point $x = 1$ (as $x = -1$ is outside the interval).

Evaluate the function $y$ at the critical point $x = 1$ .
$y(1) = \frac{1}{3}(1)^3 - 1 + 1$

Calculate $y(1)$ .
$y(1) = \frac{1}{3} - 1 + 1 = \frac{1}{3}$

Evaluate the function $y$ at the endpoints of the interval, $x = 0$ and $x = 3$ .
$y(0) = \frac{1}{3}(0)^3 - 0 + 1$

Calculate $y(0)$ .
$y(0) = 1$

Evaluate the function at the other endpoint $x = 3$ .
$y(3) = \frac{1}{3}(3)^3 - 3 + 1$

Calculate $y(3)$ .
$y(3) = \frac{1}{3}(27) - 3 + 1 = 9 - 3 + 1 = 7$

Compare the values of $y$ at the critical point and the endpoints: $y(0) = 1$ , $y(1) = \frac{1}{3}$ , $y(3) = 7$ .

Determine the absolute maximum and minimum values.
- The absolute maximum value is $7$ at $x = 3$ . - The absolute minimum value is $\frac{1}{3}$ at $x = 1$ .