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Chapter 3: Problem 32

Find the critical points of the function in the interval \([0,2 \pi] .\) Determine if each critical point is a relative maximum, a relative minimum, or neither. Use the Second-Derivative Test, when possible. Determine the points of inflection in the interval \([0,2 \pi]\). Then sketch the graph on the interval \([0,2 \pi]\) : $$ f(x)=\frac{x}{4}+\cos \frac{x}{2} $$

Short Answer

Expert verified
Critical points: \( x = \frac{\pi}{3} \) (relative maximum), \( x = \frac{5\pi}{3} \) (relative minimum). No points of inflection in \([0, 2\pi] \).

Step by step solution

01

Find the first derivative

Calculate the first derivative of \( f(x) = \frac{x}{4} + \cos \left( \frac{x}{2} \right) \). Using the chain rule, we get: \[ f'(x) = \frac{d}{dx} \left( \frac{x}{4} \right) + \frac{d}{dx} \left( \cos \left( \frac{x}{2} \right) \right) \]\[ f'(x) = \frac{1}{4} + (-\sin \left( \frac{x}{2} \right) \cdot \frac{1}{2}) \]\[ f'(x) = \frac{1}{4} - \frac{1}{2} \sin \left( \frac{x}{2} \right) \]
02

Find the critical points

Set the first derivative equal to zero and solve for \( x \): \[ \frac{1}{4} - \frac{1}{2} \sin \left( \frac{x}{2} \right) = 0 \]Isolate the sine term: \[ \sin \left( \frac{x}{2} \right) = \frac{1}{2} \]Solve for \( x \): \[ \frac{x}{2} = \frac{\pi}{6} + 2k\pi \text{ or } \frac{x}{2} = \pi - \frac{\pi}{6} + 2k\pi \]\[ x = \frac{\pi}{3} + 4k\pi \text{ or } x = \frac{5\pi}{3} + 4k\pi \]On the interval \([0, 2\pi]\), the solutions are:\[ x = \frac{\pi}{3} \text{ and } x = \frac{5\pi}{3} \]
03

Apply the Second-Derivative Test

Calculate the second derivative of \( f(x) \): \[ f''(x) = \frac{d}{dx} \left( \frac{1}{4} - \frac{1}{2} \sin \left( \frac{x}{2} \right) \right) \]\[ f''(x) = 0 - \left( \frac{1}{2} \cdot \cos \left( \frac{x}{2} \right) \cdot \frac{1}{2} \right) \]\[ f''(x) = -\frac{1}{4} \cos \left( \frac{x}{2} \right) \]Evaluate \( f''(x) \) at the critical points:For \( x = \frac{\pi}{3} \): \[ f'' \left( \frac{\pi}{3} \right) = -\frac{1}{4} \cos \left( \frac{\pi}{6} \right) = -\frac{1}{4} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{8} \]\( f'' \left( \frac{\pi}{3} \right) < 0 \), therefore, \( x = \frac{\pi}{3} \) is a relative maximum.For \( x = \frac{5\pi}{3} \): \[ f'' \left( \frac{5\pi}{3} \right) = -\frac{1}{4} \cos \left( \frac{5\pi}{6} \right) = -\frac{1}{4} \left( -\frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{8} \]\( f'' \left( \frac{5\pi}{3} \right) > 0 \), therefore, \( x = \frac{5\pi}{3} \) is a relative minimum.
04

Determine points of inflection

A point of inflection occurs where the second derivative changes sign. Set the second derivative equal to zero and solve for \( x \):\[ -\frac{1}{4} \cos \left( \frac{x}{2} \right) = 0 \] This equation is never zero, thus, there are no points of inflection in the interval \([0, 2\pi] \).
05

Sketch the graph

Based on the analysis, plot the critical points \( x = \frac{\pi}{3} \) (relative maximum) and \( x = \frac{5\pi}{3} \) (relative minimum) on the graph. The function does not have any points of inflection in the given interval. Draw the graph to reflect this behavior. Make sure the graph is periodic around these critical points within the interval \([0, 2\pi] \).

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Key Concepts

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

first derivative
The first derivative of a function provides us with the slope of the tangent line to the curve at any given point. This slope tells us how the function is increasing or decreasing at that point. For the function given in the exercise, \[ f(x) = \frac{x}{4} + \cos \left( \frac{x}{2} \right) \], the first derivative is determined using the chain rule. We calculate: \[ f'(x) = \frac{1}{4} - \frac{1}{2} \sin \left( \frac{x}{2} \right) \]. Critical points occur where this derivative is zero or undefined. By setting the first derivative to zero, \[ \frac{1}{4} - \frac{1}{2} \sin \left( \frac{x}{2} \right) = 0 \] and solving for x, we identify the potential critical points within the interval \([0, 2\pi]\).
second derivative test
The second derivative test helps in determining the concavity of the function at critical points, which allows us to classify these points as relative maxima, minima, or neither. We take the second derivative of our function: \[ f''(x) = -\frac{1}{4} \cos \left( \frac{x}{2} \right) \]. By evaluating this second derivative at our critical points \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \), we can classify them. If \( f''(x) > 0 \), it suggests a relative minimum, as the function is concave up. If \( f''(x) < 0 \), it suggests a relative maximum, as the function is concave down.
relative minimum
A relative minimum is a point where the function has a lower value than at any nearby points. We identify relative minima using the second derivative test. For our function \( f''(5\pi/3) > 0 \), thus the point \( x = 5\pi/3 \) is a relative minimum, indicating the function value at this point is less than at any nearby points. This point provides a 鈥榲alley鈥� in the graph of the function.
relative maximum
A relative maximum is a point where the function has a higher value than at any nearby points. We identify relative maxima using the second derivative test. For our function, \( f''(\pi/3) < 0 \), thus the point \( x = \pi/3 \) is a relative maximum, meaning the function value at this point is higher than at any nearby points. This point forms a 鈥榩eak鈥� on the graph of the function.
inflection points
Inflection points occur where the concavity of the function changes, meaning the second derivative changes sign. To find potential inflection points, we set the second derivative to zero and solve, \[ -\frac{1}{4} \cos \left( \frac{x}{2} \right) = 0 \]. In this exercise, there are no values for \( x \) in the interval \([0, 2\pi] \) that satisfy this equation. This implies that there are no points of inflection in the given interval. Therefore, the function does not switch from concave up to concave down (or vice versa) within this range.

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Most popular questions from this chapter

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