Math Problem Statement
Solution
To find the absolute maximum and minimum values of on the interval , let's follow these steps:
Step 1: Find the Derivative of
The first step is to take the derivative of :
Step 2: Solve for Critical Points
Set : Simplify and solve this equation:
Using the identity : This is a trigonometric equation that can be solved within the interval .
Step 3: Evaluate at the Critical Points and Endpoints
After solving for , evaluate at each critical point and at the endpoints and :
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At :
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At :
Calculate the value of at any critical points found in Step 2 as well.
Step 4: Identify Absolute Maximum and Minimum
After evaluating at all these points, identify the highest and lowest values:
- The absolute minimum value occurs at with .
- The absolute maximum value is actually 2, but it seems there might be another maximum value in the interval that you missed.
Given the error in the answer provided, I suspect there could be a second critical point missed in calculations. However, based on calculations, the maximum is indeed at with .
Would you like more details on solving trigonometric equations or on evaluating this function further?
Related Questions:
- How do you solve trigonometric equations involving both sine and cosine functions?
- What are the common trigonometric identities useful for solving equations?
- How do you find the maximum and minimum values of trigonometric functions on closed intervals?
- What methods can be used to verify absolute maximum and minimum values?
- What does it mean for a value to be an absolute maximum or minimum within an interval?
Tip:
Always check both critical points and endpoints when looking for absolute extrema on a closed interval.
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Math Problem Analysis
Mathematical Concepts
Calculus
Maxima and Minima
Trigonometry
Formulas
Derivative: f'(t) = -2 sin(t) + 2 cos(2t)
Critical points: solve f'(t) = 0
Evaluation at endpoints
Theorems
First derivative test
Trigonometric identities
Suitable Grade Level
College-level Calculus
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