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Sketch the graph of f and use your sketch to find the absolute and local maximum and minimum values of f. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x) = sin(x), −𝜋 2 ≤ x ≤ 𝜋 2 absolute maximum value absolute minimum value local maximum value(s) local minimum value(s)

Question

Sketch the graph of f and use your sketch to find the absolute and local maximum and minimum values of f. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x) = sin(x),
−𝜋 2 ≤ x ≤ 𝜋 2 absolute maximum value

absolute minimum value

local maximum value(s)

local minimum value(s)

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Solution 1

The function f(x) = sin(x) is a periodic function with a period of 2π. However, we are only considering the interval from -π/2 to π/2.

  1. Sketch the graph: The graph of f(x) = sin(x) in the interval -π/2 to π/2 starts at (0,-1), goes up to (π/2, 1), and then goes down to (π, -1).

  2. Absolute maximum value: The absolute maximum value of f(x) = sin(x) in the interval -π/2 to π/2 is 1, which occurs at x = π/2.

  3. Absolute minimum value: The absolute minimum value of f(x) = sin(x) in the interval -π/2 to π/2 is -1, which occurs at x = -π/2.

  4. Local maximum value(s): The local maximum value is the same as the absolute maximum value in this case, which is 1 at x = π/2.

  5. Local minimum value(s): The local minimum value is the same as the absolute minimum value in this case, which is -1 at x = -π/2.

So, the answers are:

Absolute maximum value: 1 Absolute minimum value: -1 Local maximum value(s): 1 Local minimum value(s): -1

This problem has been solved

Solution 2

The function f(x) = sin(x) is a periodic function with a period of 2π. However, we are only considering the interval from -π/2 to π/2.

  1. Sketch the graph: The graph of f(x) = sin(x) in the interval -π/2 to π/2 starts at (0,-1), goes up to (π/2,1), and then goes down to (π,0).

  2. Absolute maximum value: The absolute maximum value of f(x) = sin(x) in the interval -π/2 to π/2 is 1, which occurs at x = π/2.

  3. Absolute minimum value: The absolute minimum value of f(x) = sin(x) in the interval -π/2 to π/2 is -1, which occurs at x = -π/2.

  4. Local maximum value(s): The local maximum value is the same as the absolute maximum value in this case, which is 1 at x = π/2.

  5. Local minimum value(s): The local minimum value is the same as the absolute minimum value in this case, which is -1 at x = -π/2.

So, the answers are:

Absolute maximum value: 1 Absolute minimum value: -1 Local maximum value(s): 1 Local minimum value(s): -1

This problem has been solved

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