Math Problem Statement

Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 2 cos(t) + sin(2t),

0, πœ‹ 2

absolute minimum value

Correct: Your answer is correct. absolute maximum value

Solution

To find the absolute maximum and minimum values of the function f(t)=2cos⁑(t)+sin⁑(2t)f(t) = 2 \cos(t) + \sin(2t) on the interval [0,Ο€2]\left[ 0, \frac{\pi}{2} \right], we’ll follow these steps:

  1. Compute the derivative fβ€²(t)f'(t) to find the critical points.
  2. Set fβ€²(t)=0f'(t) = 0 and solve for tt within the interval.
  3. Evaluate f(t)f(t) at the critical points and endpoints t=0t = 0 and t=Ο€2t = \frac{\pi}{2}.
  4. Determine the maximum and minimum values from these evaluations.

Step 1: Compute the Derivative fβ€²(t)f'(t)

The function is given by: f(t)=2cos⁑(t)+sin⁑(2t)f(t) = 2 \cos(t) + \sin(2t)

To differentiate, we use the derivatives of cos⁑(t)\cos(t) and sin⁑(2t)\sin(2t):

fβ€²(t)=βˆ’2sin⁑(t)+2cos⁑(2t)f'(t) = -2 \sin(t) + 2 \cos(2t)

Step 2: Set fβ€²(t)=0f'(t) = 0 and Solve for tt

We need to find the values of tt that satisfy: βˆ’2sin⁑(t)+2cos⁑(2t)=0-2 \sin(t) + 2 \cos(2t) = 0

This simplifies to: sin⁑(t)=cos⁑(2t)\sin(t) = \cos(2t)

Using trigonometric identities, such as cos⁑(2t)=1βˆ’2sin⁑2(t)\cos(2t) = 1 - 2\sin^2(t), this equation can be solved to find critical points within [0,Ο€2]\left[ 0, \frac{\pi}{2} \right].


Step 3: Evaluate f(t)f(t) at Critical Points and Endpoints

Once we have critical points, we evaluate f(t)f(t) at these points as well as at t=0t = 0 and t=Ο€2t = \frac{\pi}{2}.

  1. f(0)=2cos⁑(0)+sin⁑(0)=2f(0) = 2 \cos(0) + \sin(0) = 2
  2. f(Ο€2)=2cos⁑(Ο€2)+sin⁑(Ο€)=0f\left(\frac{\pi}{2}\right) = 2 \cos\left(\frac{\pi}{2}\right) + \sin(\pi) = 0

Step 4: Determine Maximum and Minimum Values

By comparing values from critical points and endpoints, we find the absolute maximum and absolute minimum values over the interval.

Would you like me to go further with specific critical points or evaluate the function with detailed calculations?


Here are related questions you might explore:

  1. How do critical points affect the outcome of maximum and minimum values?
  2. What is the importance of checking endpoints when finding absolute extrema?
  3. How does the derivative test help in finding extrema on closed intervals?
  4. What are alternative methods for verifying critical points in trigonometric functions?
  5. How does trigonometric identity simplification help in solving equations?

Tip: Always consider both endpoints and critical points within a given interval to ensure you have the absolute extrema for a function.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometric Functions
Optimization

Formulas

Derivative of f(t) to find critical points
Evaluating f(t) at critical points and endpoints

Theorems

Extreme Value Theorem

Suitable Grade Level

Grades 11-12, Calculus I level