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Chapter 4: Problem 19

Use the First Derivative Test to determine the relative extreme values (if any) of the function. $$ k(x)=\cos x+\frac{1}{2} x $$

Short Answer

Expert verified
Relative minimum at \( x = \frac{\pi}{6} \) and relative maximum at \( x = \frac{5\pi}{6} \).

Step by step solution

01

Find the derivative of the function

First, we need to find the derivative of the function \( k(x) = \cos x + \frac{1}{2}x \). The derivative of \( \cos x \) is \( -\sin x \) and the derivative of \( \frac{1}{2}x \) is \( \frac{1}{2} \). Combining these, we get the derivative: \( k'(x) = -\sin x + \frac{1}{2} \).
02

Set the derivative to zero and solve for x

We set the derivative equal to zero to find critical points: \( -\sin x + \frac{1}{2} = 0 \). Solving for \( \sin x \), we get \( \sin x = \frac{1}{2} \). This occurs at \( x = \frac{\pi}{6} + 2n\pi \) and \( x = \frac{5\pi}{6} + 2n\pi \) where \( n \) is an integer.
03

Use the First Derivative Test

We analyze the intervals around the critical points. Choose test values within the intervals to determine the sign of \( k'(x) \). For example:- For \( x \) slightly less than \( \frac{\pi}{6} \), \( k'(x) < 0 \) (since \( \sin x < \frac{1}{2} \)). - For \( x \) slightly greater than \( \frac{\pi}{6} \), \( k'(x) > 0 \). - For \( x \) slightly less than \( \frac{5\pi}{6} \), \( k'(x) > 0 \). - For \( x \) slightly greater than \( \frac{5\pi}{6} \), \( k'(x) < 0 \). Thus, \( x = \frac{\pi}{6} \) is a relative minimum and \( x = \frac{5\pi}{6} \) is a relative maximum.
04

Verify the extrema values

Substitute the critical points back into the original function to find the relative extrema values. For \( x = \frac{\pi}{6} \), \( k(\frac{\pi}{6}) = \cos(\frac{\pi}{6}) + \frac{1}{2}\cdot\frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{\pi}{12} \). For \( x = \frac{5\pi}{6} \), \( k(\frac{5\pi}{6}) = \cos(\frac{5\pi}{6}) + \frac{1}{2} \cdot \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{5\pi}{12} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points are key to understanding where a function's behavior changes. They typically occur where the derivative of a function is zero or undefined. For the function \( k(x) = \cos x + \frac{1}{2}x \), finding those points requires setting its derivative \( k'(x) = -\sin x + \frac{1}{2} \) equal to zero, as seen in the exercise.
  • Critical points indicate where a possible peak (maximum) or trough (minimum) might be.
  • They are not always points of extreme values but must be tested further.
Critical points for trigonometric functions such as cosine often repeat periodically due to their cyclical nature. This means more than one solution could exist within different intervals.
Relative Extrema
Relative extrema are the peaks and valleys within a specific region of a graph. They are determined through the First Derivative Test after identifying critical points. For our function \( k(x) \), these points were found by evaluating \( k'(x) \) around the critical values. A change in sign of the derivative indicates a relative extremum:
  • A **relative minimum** occurs if the derivative changes from negative to positive.
  • A **relative maximum** happens when the derivative shifts from positive to negative.
In our example, \( x = \frac{\pi}{6} \) offers a relative minimum because \( k'(x) \) transitions from negative to positive. On the other hand, \( x = \frac{5\pi}{6} \) is a relative maximum as the derivative changes from positive to negative.
Trigonometric Functions
Trigonometric functions, like \( \cos x \), have unique properties which affect their derivatives and subsequent analysis. These functions exhibit periodic behavior, repeating their values in regular intervals, such as \( 2\pi \) for cosine.
  • The derivative of a trigonometric function can lead to periodic solutions, resulting in multiple potential critical points.
  • Understanding the trigonometric identities and derivatives is crucial for solving related calculus problems effectively.
In the given task, identifying angles where \( \sin x = \frac{1}{2} \) requires knowledge of common angle values from the unit circle, namely \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), among others.
Derivative Computation
Calculating the derivative correctly is foundational in applying the First Derivative Test and overall calculus problem-solving. Here, the function is composed of a trigonometric part and a linear part.
  • The derivative of \( \cos x \) is \( -\sin x \), reflecting the sinusoidal nature of these functions.
  • The derivative of a linear function, such as \( \frac{1}{2}x \), is simply a constant: \( \frac{1}{2} \).
Combining these results into \( k'(x) = -\sin x + \frac{1}{2} \) gives a new function to analyze. Correct computation allows for precise determination of where the original function is increasing or decreasing, which is essential for finding critical points and relative extrema. Ensure accuracy by applying differentiation rules correctly and confirming results with interval testing.

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