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Chapter 3: Problem 67

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 \((-\infty, \infty)\). $$ f(x)=\tan x-2 \sec x, \quad(\pi / 2,3 \pi / 2) $$

Short Answer

Expert verified
Critical Points are \( x = \frac{7\pi}{6}\). Evaluate at critical points and boundaries for \(f(x)\).

Step by step solution

01

- Understand the Function and Interval

The function is given as \(f(x) = \tan x - 2 \sec x\), and the interval is \((\pi/2, 3\pi/2)\). This interval is important because it will dictate where to look for extrema.
02

- Find the First Derivative

Calculate the first derivative of the function to find the critical points. Use the derivatives \(\frac{d}{dx} [\tan(x)] = \sec^2(x)\) and \(\frac{d}{dx} [\sec(x)] = \sec(x) \tan(x)\). Thus, \[f'(x) = \sec^2(x) - 2 \sec(x) \tan(x)\]
03

- Set the Derivative Equal to Zero

Solve \(f'(x) = 0\) to find critical points: \[\sec^2(x) - 2 \sec(x) \tan(x) = 0\]Factor out \(\sec(x)\): \[\sec(x) (\sec(x) - 2 \tan(x)) = 0\]
04

- Solve for Critical Points

Since \(\sec(x) eq 0\), solve for \(\sec(x) = 2 \tan(x)\): \[\sec(x) = \frac{1}{\cos(x)}\]\[\tan(x) = \frac{\sin(x)}{\cos(x)}\]Thus, \[\frac{1}{\cos(x)} = 2 \cdot \frac{\sin(x)}{\cos(x)}\]\[1 = 2 \sin(x)\]\[\sin(x) = \frac{1}{2}\]
05

- Identify Valid Solutions within the Interval

Find the angles in the interval \((\pi/2, 3\pi/2)\) where \(\sin(x) = \frac{1}{2}\). Since \(\sin(x) = \frac{1}{2}\) at \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\), but these must be converted into terms within the given interval: \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}; x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\). Check if these angles lie within \((\pi/2, 3\pi/2)\) for valid critical points.
06

- Evaluate the Function at Critical Points

Evaluate \(f(x)\) at the critical points. First point is \(x = \frac{7\pi}{6}\):\[f\left(\frac{7\pi}{6}\right) = \tan\left(\frac{7\pi}{6}\right) - 2 \sec\left(\frac{7\pi}{6}\right)\]Since \(\tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) = \sqrt{3}\), and \(\sec\left(\frac{7\pi}{6}\right) = -\frac{2}{\sqrt{3}}\). Similar evaluation must be done for next critical point.
07

- Evaluate the Function at Endpoints of the Interval

Evaluate the function at the boundaries of the interval: \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\). Calculate these values to find the absolute extrema values.

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Key Concepts

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

absolute maximum
In calculus, finding the absolute maximum of a function means determining the highest value that the function attains over a given interval. This value is the peak of the function within the specified range. When looking for absolute maxima, it's important to evaluate the function at critical points and endpoints within the interval. In the provided exercise, the function is \(f(x) = \tan x - 2 \sec x\), and we evaluated it over the interval \((\pi/2, 3\pi/2)\). For any function, the absolute maximum helps in understanding the highest point the function reaches in real-world contexts. This is crucial in life sciences where it might represent the maximum population size, concentration level, or any other biological peak.
absolute minimum
Finding the absolute minimum of a function involves identifying the lowest value the function reaches over an interval. This value is the trough of the function within the specified range. Like the absolute maximum, the absolute minimum is found by evaluating the function at its critical points and at the boundaries of the interval. In this exercise, we checked the function \(f(x) = \tan x - 2 \sec x\) within the interval \((\pi/2, 3\pi/2)\) to locate the absolute minimum. The absolute minimum is key in life sciences for understanding the least amount or deficiency in a system, such as the lowest concentration of a substance or minimum population density.
critical points
Critical points of a function are the values of \(x\) where the first derivative of the function is zero or undefined. These points are crucial because they are potential locations where the function reaches local or absolute maxima or minima. To find the critical points in the provided problem, we calculated the derivative of \(f(x) = \tan x - 2 \sec x\) and set it equal to zero:

\[f'(x) = \sec^2(x) - 2 \sec(x) \tan(x) = 0\]
Solving this helps us locate the critical points within the interval \((\pi/2, 3\pi/2)\). Understanding critical points is essential as they pinpoint where significant changes, such as turning points or periods of equilibrium, occur in biological systems.
derivative
A derivative represents the rate at which a function is changing at any given point. It gives the slope of the function at a particular point, indicating whether the function is increasing or decreasing. Calculating the first derivative is the key to finding critical points. For the function \(f(x) = \tan x - 2 \sec x\), the first derivative is:

\[f'(x) = \sec^2(x) - 2 \sec(x) \tan(x)\]

This calculation uses the derivatives of \(\tan(x)\) and \(\sec(x)\). By setting the first derivative to zero, we identify where the function's rate of change is zero, which indicates possible maxima, minima, or inflection points. This serves as a crucial step in analyzing biological trends, like changes in population growth rates or concentrations over time.
interval evaluation
Interval evaluation involves examining the function over a specific range of values. In this process, we look for the function's behavior at both the critical points and the endpoints of the interval. For the function \(f(x) = \tan x - 2 \sec x\) over the interval \((\pi/2, 3\pi/2)\), we evaluate the function at each critical point and at the interval's boundaries:
  • First critical point: \(x = 7\pi/6\)
  • Second critical point: evaluated similarly
These evaluations help determine the absolute maximum and minimum values over the interval. Interval evaluation is vital in life sciences to determine the optimal or extreme values of a phenomenon within a specified range, such as seasonal population peaks or dips.

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Most popular questions from this chapter

Graph the function. $$ f(x)=\left|\frac{1}{x}-2\right| $$

Differentiate implicily to find \(d y / d x\). $$ x+y=\sin (\sqrt{y-x}) $$

Use the indicated choice of \(x_{1}\) and Newton's method to solve the given equation. \(2 x-\sin x=\cos \left(x^{2}\right) ; x_{1}=\pi / 4\)

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.

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 \((-\infty, \infty)\). $$ f(x)=x^{2 / 3} ; \quad[-1,1] $$
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