Math / Calculus

QuestionThe following function is continuous and defined on the closed interval. It therefore satisfies the assumptions of the extreme-value theorem. With the help of technology, graph each function a locate its global extrema. $f(x)=-x^{2}+1,-2 \leq x \leq 2$
Graph the function. Choose the correct answer below. A. B. c. D.
Find the absolute maximum value. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The absolute maximum value 1 occurs at $x=0$ . (Simplify your answers. Use a comma to separate answers as needed.) B. There is no absolute maximum.
Find the absolute minimum value. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The absolute minimum value $\square$ occurs at $x=$ $\square$ $\square$ . (Simplify your answers. Use a comma to separate answers as needed.) B. There is no absolute minimum.

Studdy Solution

STEP 1

1. The function $f(x) = -x^2 + 1$ is continuous on the interval $[-2, 2]$ .
2. The extreme-value theorem applies, ensuring both a maximum and minimum exist on this interval.
3. We need to find the global extrema of the function within the given interval.

STEP 2

1. Analyze the function to understand its shape and behavior.
2. Determine the critical points within the interval.
3. Evaluate the function at the critical points and endpoints to find the extrema.
4. Identify the absolute maximum and minimum values.

STEP 3

The function $f(x) = -x^2 + 1$ is a downward-opening parabola because the coefficient of $x^2$ is negative.

STEP 4

Find the derivative of $f(x)$ to locate critical points: $f'(x) = -2x$
Set the derivative equal to zero to find critical points: $-2x = 0$ $x = 0$

STEP 5

Evaluate the function at the critical point and the endpoints of the interval: $f(-2) = -(-2)^2 + 1 = -4 + 1 = -3$ $f(0) = -(0)^2 + 1 = 1$ $f(2) = -(2)^2 + 1 = -4 + 1 = -3$

STEP 6

Compare the values obtained: - $f(-2) = -3$ - $f(0) = 1$ - $f(2) = -3$
The absolute maximum value is $1$ at $x = 0$ . The absolute minimum value is $-3$ at $x = -2$ and $x = 2$ .
The correct answers are: - Absolute maximum: A. The absolute maximum value 1 occurs at $x = 0$ . - Absolute minimum: A. The absolute minimum value $-3$ occurs at $x = -2, 2$ .

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Studdy Solution

STEP 1

1. The function $f(x) = -x^2 + 1$ is continuous on the interval $[-2, 2]$ .
2. The extreme-value theorem applies, ensuring both a maximum and minimum exist on this interval.
3. We need to find the global extrema of the function within the given interval.

STEP 2

1. Analyze the function to understand its shape and behavior.
2. Determine the critical points within the interval.
3. Evaluate the function at the critical points and endpoints to find the extrema.
4. Identify the absolute maximum and minimum values.

STEP 3

The function $f(x) = -x^2 + 1$ is a downward-opening parabola because the coefficient of $x^2$ is negative.

STEP 4

Find the derivative of $f(x)$ to locate critical points: $f'(x) = -2x$
Set the derivative equal to zero to find critical points: $-2x = 0$ $x = 0$

STEP 5

Evaluate the function at the critical point and the endpoints of the interval: $f(-2) = -(-2)^2 + 1 = -4 + 1 = -3$ $f(0) = -(0)^2 + 1 = 1$ $f(2) = -(2)^2 + 1 = -4 + 1 = -3$

STEP 6

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Studdy Solution

STEP 1

1. The function f(x)=−x2+1 f(x) = -x^2 + 1 f(x)=−x2+1 is continuous on the interval [−2,2][-2, 2][−2,2].2. The extreme-value theorem applies, ensuring both a maximum and minimum exist on this interval.3. We need to find the global extrema of the function within the given interval.

STEP 2

1. Analyze the function to understand its shape and behavior.2. Determine the critical points within the interval.3. Evaluate the function at the critical points and endpoints to find the extrema.4. Identify the absolute maximum and minimum values.

STEP 3

The function f(x)=−x2+1 f(x) = -x^2 + 1 f(x)=−x2+1 is a downward-opening parabola because the coefficient of x2 x^2 x2 is negative.

STEP 4

Find the derivative of f(x) f(x) f(x) to locate critical points: f′(x)=−2x f'(x) = -2x f′(x)=−2xSet the derivative equal to zero to find critical points: −2x=0 -2x = 0 −2x=0 x=0 x = 0 x=0

STEP 5

STEP 6

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1. The function $f(x) = -x^2 + 1$ is continuous on the interval $[-2, 2]$ .
2. The extreme-value theorem applies, ensuring both a maximum and minimum exist on this interval.
3. We need to find the global extrema of the function within the given interval.

1. Analyze the function to understand its shape and behavior.
2. Determine the critical points within the interval.
3. Evaluate the function at the critical points and endpoints to find the extrema.
4. Identify the absolute maximum and minimum values.

The function $f(x) = -x^2 + 1$ is a downward-opening parabola because the coefficient of $x^2$ is negative.

Find the derivative of $f(x)$ to locate critical points: $f'(x) = -2x$
Set the derivative equal to zero to find critical points: $-2x = 0$ $x = 0$