Math Problem Statement

Find the absolute extrema of f(x) =sin x on [0, π]

Solution

To find the absolute extrema (the maximum and minimum values) of the function f(x)=sin(x)f(x) = \sin(x) on the interval [0,π][0, \pi], we need to follow these steps:

Step 1: Check the critical points

Critical points occur where the derivative of the function is either zero or undefined.

  1. First, find the derivative of f(x)f(x): f(x)=ddx(sin(x))=cos(x)f'(x) = \frac{d}{dx} (\sin(x)) = \cos(x)

  2. Set the derivative equal to zero to find critical points: cos(x)=0\cos(x) = 0

  3. Solve for xx within the interval [0,π][0, \pi]: cos(x)=0x=π2\cos(x) = 0 \quad \Rightarrow \quad x = \frac{\pi}{2} So, x=π2x = \frac{\pi}{2} is a critical point within the interval [0,π][0, \pi].

Step 2: Evaluate the function at the endpoints of the interval and at the critical points

Now, we evaluate f(x)=sin(x)f(x) = \sin(x) at the endpoints of the interval [0,π][0, \pi] and at the critical point x=π2x = \frac{\pi}{2}.

  • At x=0x = 0: f(0)=sin(0)=0f(0) = \sin(0) = 0
  • At x=πx = \pi: f(π)=sin(π)=0f(\pi) = \sin(\pi) = 0
  • At x=π2x = \frac{\pi}{2} (the critical point): f(π2)=sin(π2)=1f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1

Step 3: Identify the absolute maximum and minimum

From the evaluations above, we have:

  • f(0)=0f(0) = 0
  • f(π)=0f(\pi) = 0
  • f(π2)=1f\left(\frac{\pi}{2}\right) = 1

The absolute maximum value of f(x)f(x) on the interval [0,π][0, \pi] is 11, which occurs at x=π2x = \frac{\pi}{2}, and the absolute minimum value is 00, which occurs at both x=0x = 0 and x=πx = \pi.

Final Answer:

  • The absolute maximum value of f(x)=sin(x)f(x) = \sin(x) on [0,π][0, \pi] is 11 at x=π2x = \frac{\pi}{2}.
  • The absolute minimum value of f(x)=sin(x)f(x) = \sin(x) on [0,π][0, \pi] is 00 at x=0x = 0 and x=πx = \pi.

Would you like to go over the steps in more detail, or do you have any specific questions?

Related Questions:

  1. How does the derivative f(x)=cos(x)f'(x) = \cos(x) behave over the interval [0,π][0, \pi]?
  2. What would the extrema be for f(x)=cos(x)f(x) = \cos(x) on the same interval?
  3. How do absolute extrema differ from local extrema?
  4. What would happen if the interval was [0,2π][0, 2\pi]?
  5. Can we apply the same method to other trigonometric functions, like tan(x)\tan(x)?

Tip:

Remember that the critical points are where the derivative equals zero or is undefined. For continuous functions, such as sine and cosine, the extrema can only occur either at these critical points or the boundaries of the interval.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometry
Critical Points
Extrema

Formulas

f'(x) = cos(x)
f(x) = sin(x)

Theorems

First derivative test for extrema
Absolute extrema on a closed interval

Suitable Grade Level

Grades 11-12