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Chapter 3: Problem 36

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 \([0,2 \pi]\) : $$ \int(x)=\sin x-\sqrt{3} \cos x $$

Short Answer

Expert verified
Critical points at \( x = \frac{5\pi}{6} \) (minimum) and \( x = \frac{11\pi}{6} \) (inflection). No inflection points found elsewhere.

Step by step solution

01

Find the First Derivative

Compute the first derivative of the function \( f(x) = \sin(x) - \sqrt{3} \cos(x) \). Using differentiation rules, you get the following: \( f'(x) = \cos(x) + \sqrt{3} \sin(x) \).
02

Set the First Derivative to Zero

To find the critical points, solve \( f'(x) = 0 \). This gives: \( \cos(x) + \sqrt{3} \sin(x) = 0 \). Rearranging terms gives: \( \cos(x) = - \sqrt{3} \sin(x). \)
03

Solve for \( x \)

Divide both sides by \( \cos(x) \) (assuming \(\cos(x) \eq 0 \)), then use the identity \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) to get: \( \tan(x) = -\frac{1}{\sqrt{3}} \). Thus, \( x = \frac{5\pi}{6} \) and \( x = \frac{11\pi}{6} \) in the interval \([0, 2\pi] \).
04

Second Derivative to Confirm Nature of Critical Points

Find the second derivative with respect to \( x \) for \( f(x) = \sin(x) - \sqrt{3} \cos(x) \). First derivative: \( f'(x) = \cos(x) + \sqrt{3} \sin(x) \). Second derivative: \( f''(x) = -\sin(x) + \sqrt{3} \cos(x) \). Evaluate \( f''(x) \) at the critical points. \( f''(\frac{5\pi}{6}) = -\sin(\frac{5\pi}{6}) + \sqrt{3} \cos(\frac{5\pi}{6}) = -\frac{1}{2}(+) + \sqrt{3}(+)-\frac{\sqrt{3}}{2} = -\frac{1}{2} + \frac{3}{2} = 1 \) => \( x = \frac{5\pi}{6} \) is a relative minimum point. Similarly, \( f''(\frac{11\pi}{6}) = -\sin(\frac{11\pi}{6}) + \sqrt{3} \cos(\frac{11\pi}{6}) = - \frac{-1}{2} - \frac{\sqrt{3}}{2}(.77) \) which means it's negative, which purely equals -1/2 Total due to inflection point(11蟺6).
05

Identify Points of Inflection

A point of inflection occurs where the second derivative changes sign. Check where \( f''(x) \) changes sign. Solve \( f''(x) = 0 \).
06

Sketch the Graph

Use the critical points and inflection points to draw the graph of the function \( f(x) = \sin(x) - \sqrt{3} \cos(x) \) over the interval \([0, 2\pi] \). Mark the critical points \( x = \frac{5\pi}{6} \) and \( x = \frac{11\pi}{6} \), and annotate their nature as determined. Also, label any points of inflection found.

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Key Concepts

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

First Derivative
To understand critical points, we first need the first derivative of the function. The first derivative helps identify where the slope of the function is zero. For the function given, \( f(x) = \sin(x) - \sqrt{3} \cos(x) \), applying basic differentiation rules gives us: \( f'(x) = \cos(x) + \sqrt{3} \sin(x) \). Finding this derivative is crucial since critical points occur where this first derivative equals zero. Hence, we set \( f'(x) = 0 \) and solve for \( x \). This step is foundational for identifying where the function has potential maxima, minima, or saddle points.
Second Derivative
The second derivative offers insight into the concavity of the function. For our function, the second derivative is given by: \( f''(x) = -\sin(x) + \sqrt{3} \cos(x) \). The second derivative tells us how the slope of the function is changing. It's instrumental in distinguishing between relative maxima, minima, and points of inflection. When evaluating the second derivative at critical points, we can determine the curvature: if \( f''(x) > 0 \), the function is concave up, hinting at a relative minimum. Conversely, if \( f''(x) < 0 \), the function is concave down, indicating a relative maximum.
Relative Maximum
A relative maximum is where the function reaches a peak within a specific interval. To identify this, solve \( f''(x) \) for its sign at critical points. However, from our calculations: at \( x = \frac{5\pi}{6} \), \( f''(x) = 1 \), which is positive. Thus, it鈥檚 a relative minimum. So we can elaborate that if \( f''(x) \) were negative at a critical point, it would signify a relative maximum. For our function, the evaluation for \( x = \frac{11\pi}{6} \) also doesn鈥檛 yield a relative maximum.
Relative Minimum
A relative minimum occurs where the function dips to a valley within a given range. Using the second derivative test: our critical point \( x = \frac{5\pi}{6} \) turns out to be a relative minimum due to \( f''(\frac{5\pi}{6}) = 1 \). This means the function changes from decreasing to increasing, creating a valley at this point. Remember, a relative minimum is confirmed when \( f''(x) \) at a critical point is positive.
Points of Inflection
Points of inflection are where the function changes concavity. This happens when \( f''(x) = 0 \) and changes sign. For the function \( \sin(x) - \sqrt{3} \cos(x) \), identify where \( f''(x) = -\sin(x) + \sqrt{3} \cos(x) = 0 \). Solving this equation will highlight our inflection points. These points are essential as they signify a shift from concave up to concave down or vice versa. Finding such points helps in sketching the precise shape of the graph.

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Most popular questions from this chapter

Find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. When no interval is specified, use the real line \((-\infty, \infty)\). $$ f(x)=\frac{\sin x}{1+\sin x} ; \quad(-\pi / 2,3 \pi / 2) $$

Differentiate implicity to find \(d y / d x\) and \(d^{2} y / d x^{2}\). $$ y^{2}-x y+x^{2}=5 $$

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When we speak, the position of the tongue changes the shapes of the mouth and throat cavity, thereby changing the sound that is generated. For example, to pronounce the first vowel of the word father, the mouth is much wider than the throat cavity. For many vowel sounds, the vocal tract may be modeled by two adjoining tubes, with one end closed (the glottis) and the other end open (the lips). We denote by \(A_{m}\) and \(L_{m}\) the cross-sectional area and length of the mouth, and we denote by \(A_{t}\) and \(L_{t}\) the cross-sectional area and length of the throat cavity. The shape of the vocal tract tends to promote certain sound frequencies. \({ }^{15}\) Let \(c\) be the speed of sound. If \(\int\) is a frequency promoted by the vocal tract (in Hertz) and we let \(x=2 \pi \int / c\), then \(x\) is a solution of $$ \frac{1}{A_{m}} \tan \left(L_{m} x\right)-\frac{1}{A_{t}} \cot \left(L_{t} x\right)=0 $$ a) For a certain speaker of the first vowel of the word father; \(A_{m}=10 A_{t}\) (that is, the mouth opening is 10 times larger than the throat opening),\(L_{m}=8 \mathrm{~cm}\), and \(L_{t}=9.7 \mathrm{~cm} .^{16}\) Use this information to show that $$ \tan (8 x)-10 \cot (9.7 x)=0 $$ a) For a certain speaker of the first vowel of the word father; \(A_{m}=10 A_{t}\) (that is, the mouth opening is 10 times larger than the throat opening),\(L_{m}=8 \mathrm{~cm}\), and \(L_{t}=9.7 \mathrm{~cm} .^{16}\) Use this information to show that $$ \tan (8 x)-10 \cot (9.7 x)=0 $$ b) Use a grapher to sketch the graph of \(y=\tan (8 x)-10 \cot (9.7 x)\) using the graphing window \([0,0.5,-2,2] .\) From the graph, estimate the first three \(x\) -intercepts. c) Use Newton's method to find the first three solutions. Use \(x_{1}=0.1, x_{1}=0.2\), and \(x_{1}=0.45\) as your three starting points. d) The speed of sound is approximately \(35,400 \mathrm{~cm} / \mathrm{s}\). Use this value for \(c\) and the relationship \(x=2 \pi f / c\) to find the first three natural frequencies of the speaker's vocal tract.
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