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Chapter 3: Problem 56

Graphing with inverse trigonometric functions a. Graph the function \(f(x)=\frac{\tan ^{-1} x}{x^{2}+1}\) b. Compute and graph \(f^{\prime}\) and determine (perhaps approximately) the points at which \(f^{\prime}(x)=0\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has a horizontal tangent line.

Short Answer

Expert verified
In this problem, we were given the function \(f(x) = \frac{\tan^{-1}(x)}{x^2 + 1}\) and asked to find its derivative, determine the points at which the derivative is zero, and verify that those points correspond to horizontal tangent lines of the original function. After applying the quotient rule and simplifying the expression, we found \(f'(x) = \frac{1 - 2x\tan^{-1}(x)}{(x^2 + 1)^2}\). To find the zeros of the derivative, we need to solve the equation \(1 - 2x\tan^{-1}(x) = 0\). We noticed that this equation does not have a simple algebraic solution and requires numerical methods or graphical means to approximate the zeros. Finally, we concluded that the zeros of \(f'(x)\) correspond to points where the function has horizontal tangent lines by visually checking the graph of \(f(x)\) at those points.

Step by step solution

01

Graph the function \(f(x) = \frac{\tan^{-1}(x)}{x^2 + 1}\)

To graph the function \(f(x) = \frac{\tan^{-1}(x)}{x^2 + 1}\), you would typically use software (like a graphing calculator or a computer program) to generate the plot. The graph will help us visualize the behavior of the function.
02

Compute the derivative of the function and graph it

To find the derivative of the function \(f(x) = \frac{\tan^{-1}(x)}{x^2 + 1}\), we will apply the quotient rule. The quotient rule states that if \(u\) and \(v\) are functions of \(x,\) then the derivative of \(\frac{u}{v}\) is given by: $$\frac{d}{dx}\left( \frac{u}{v}\right) = \frac{vu' - uv'}{v^2}$$ In our case, we have: \(u = \tan^{-1}(x) \implies u' = \frac{1}{1+x^2}\) \(v = x^2 +1 \implies v' = 2x\) Now applying the quotient rule: $$f'(x) = \frac{(x^2 + 1) (\frac{1}{1+x^2}) - (\tan ^{-1}(x))(2x)}{(x^2 + 1)^2}$$ Simplifying it, we have: $$f'(x) = \frac{1 - 2x\tan^{-1}(x)}{(x^2 + 1)^2}$$ To graph \(f'(x)\), you would again use graphing software, which will allow us to study the behavior of the derivative.
03

Find zeros of \(f'(x)\)

To find the zeros of \(f'(x)\), we need to solve the equation: $$f'(x) = 0 \implies \frac{1 - 2x\tan^{-1}(x)}{(x^2 + 1)^2} = 0$$ Since the denominator cannot be equal to zero, we only need to focus on the numerator: $$1 - 2x\tan^{-1}(x)= 0$$ $$2x\tan^{-1}(x) = 1$$ This equation might not have a simple algebraic solution, and numerical methods such as the bisection method, Newton's method, or the secant method can be employed to find the zeros, or we can approximate using graphical means.
04

Verify that the zeros of \(f'(x)\) correspond to points at which \(f\) has a horizontal tangent line.

The zeros of \(f'(x)\) are the points where the derivative is zero, and a zero derivative means the function has a horizontal tangent line at those points. We can visually check this by analyzing the graph of \(f(x)\) at the points where \(f'(x) = 0\). If the tangent line appears to be horizontal at those points, our assertion is correct.

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Key Concepts

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

Quotient Rule
When we want to find the derivative of a function that is a quotient of two other functions, we use the Quotient Rule. It is a method that helps us when we have a function like \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are functions of \( x \). The formula for the Quotient Rule is:
  • \[ \frac{d}{dx}\left( \frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \]
To apply this rule, follow these steps:
  • Identify \( u \) and \( v \).
  • Calculate the derivatives \( u' \) and \( v' \).
  • Substitute these values into the Quotient Rule formula.
  • Simplify the expression to find the derivative.
Applying the Quotient Rule allows us to break down complex functions into simpler parts, making it easier to analyze their behavior.
Derivative Analysis
Derivative analysis involves finding and understanding the derivative of a function. The derivative represents the rate of change of the function concerning its independent variable, often \( x \). For the function \( f(x) = \frac{\tan^{-1}(x)}{x^2 + 1} \), the derivative tells us how the function's value changes as \( x \) changes.In our particular problem, after applying the Quotient Rule, we found that:
  • \[ f'(x) = \frac{1 - 2x \tan^{-1}(x)}{(x^2 + 1)^2} \]
By analyzing this derivative, we can gather insights on:
  • Where the function is increasing or decreasing.
  • Critical points where changes in behavior occur, particularly where the derivative equals zero.
These critical points lead us to further investigate the nature of horizontal tangents, where the slope of the function becomes zero, indicating a potential maximum, minimum, or inflection point.
Graphing Functions
Graphing functions is crucial to visually understand the behavior of mathematical models. When working with inverse trigonometric functions like \( f(x) = \frac{\tan^{-1}(x)}{x^2 + 1} \), using graphing software can be incredibly helpful. Creating a graph provides insights into:
  • The overall shape of the function.
  • Points where the function attains certain key characteristics.
  • The behavior of the function at various intervals of \( x \).
For this function, the graph will show us where it increases and decreases, and help us find points of interest such as intercepts and asymptotic behavior. Additionally, once we plot \( f'(x) \), we see where the slopes of \( f(x) \) change direction, indicative of horizontal tangents.
Horizontal Tangent
A horizontal tangent line to a function at a particular point means the slope of the function at that point is zero. In other words, the function is not increasing or decreasing at that exact point.When examining the function \( f(x) = \frac{\tan^{-1}(x)}{x^2 + 1} \), we find the derivative:
  • \[ f'(x) = \frac{1 - 2x \tan^{-1}(x)}{(x^2 + 1)^2} \]
Setting this equation to zero helps us find where horizontal tangents occur:
  • \[ 1 - 2x\tan^{-1}(x) = 0 \]
Solving this tells us the specific \( x \)-values where the function \( f(x) \) has a flat or a stationary point. At these points, there is a possibility of a local maximum or minimum, and it is crucial for understanding the behavior of the function overall.

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Most popular questions from this chapter

Electrostatic force The magnitude of the electrostatic force between two point charges \(Q\) and \(q\) of the same sign is given by \(F(x)=\frac{k Q q}{x^{2}},\) where \(x\) is the distance (measured in meters) between the charges and \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\) is a physical constant (C stands for coulomb, the unit of charge; N stands for newton, the unit of force). a. Find the instantaneous rate of change of the force with respect to the distance between the charges. b. For two identical charges with \(Q=q=1 \mathrm{C},\) what is the instantaneous rate of change of the force at a separation of \(x=0.001 \mathrm{m} ?\) c. Does the magnitude of the instantaneous rate of change of the force increase or decrease with the separation? Explain.

Graphing \(f\) and \(f^{\prime}\) a. Graph \(f\) with a graphing utility. b. Compute and graph \(f^{\prime}\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has \(a\) horizontal tangent line. $$f(x)=(x-1) \sin ^{-1} x \text { on }[-1,1]$$

Use a trigonometric identity to show that the derivatives of the inverse cotangent and inverse cosecant differ from the derivatives of the inverse tangent and inverse secant, respectively, by a multiplicative factor of -1

Suppose \(y=L(x)=a x+b\) (with \(a \neq 0\) ) is the equation of the line tangent to the graph of a one-to-one function \(f\) at \(\left(x_{0}, y_{0}\right) .\) Also, suppose that \(y=M(x)=c x+d\) is the equation of the line tangent to the graph of \(f^{-1}\) at \(\left(y_{0}, x_{0}\right)\) a. Express \(a\) and \(b\) in terms of \(x_{0}\) and \(y_{0}\) b. Express \(c\) in terms of \(a,\) and \(d\) in terms of \(a, x_{0},\) and \(y_{0}\) c. Prove that \(L^{-1}(x)=M(x)\)

An airliner passes over an airport at noon traveling \(500 \mathrm{mi} / \mathrm{hr}\) due west. At \(1: 00 \mathrm{p} . \mathrm{m} .,\) another airliner passes over the same airport at the same elevation traveling due north at \(550 \mathrm{mi} / \mathrm{hr} .\) Assuming both airliners maintain their (equal) elevations, how fast is the distance between them changing at 2: 30 p.m.?
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