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Chapter 7: Problem 134

Use your graphing utility. Graph \(f(x)=\tan ^{-1} x\) together with its first two derivatives. Comment on the behavior of \(f\) and the shape of its graph in relation to the signs and values of \(f^{\prime}\) and \(f^{\prime \prime}\).

Short Answer

Expert verified
\( f(x) = \tan^{-1}(x) \) increases with decreasing slope; \( f'(x) > 0 \) and \( f''(x) \) indicates concavity changes at \( x = 0 \).

Step by step solution

01

Understanding the Function

The function given is the inverse tangent function \( f(x) = \tan^{-1}(x) \). The inverse tangent function, also known as the arctan function, outputs angles for given input values, and its range is \((-\pi/2, \pi/2)\).
02

Compute the First Derivative

The first derivative of the function \( f(x) = \tan^{-1}(x) \) is \( f'(x) = \frac{d}{dx}\tan^{-1}(x) = \frac{1}{1+x^2} \). This derivative gives us the slope of the tangent line to the function at any given point.
03

Compute the Second Derivative

The second derivative involves differentiating \( f'(x) = \frac{1}{1+x^2} \). Using the chain rule, we find \( f''(x) = \frac{-2x}{(1+x^2)^2} \). This helps us understand the concavity of the function.
04

Graph the Functions

Use a graphing utility to plot \( f(x) = \tan^{-1}(x) \), \( f'(x) = \frac{1}{1+x^2} \), and \( f''(x) = \frac{-2x}{(1+x^2)^2} \). Observe how each graph relates to the original function. Note that \( f'(x) \) is always positive but decreases as \( x \) moves away from zero, indicating \( f(x) \) is increasing and flattening out. The second derivative \( f''(x) \) indicates \( f(x) \) is concave downward when \( x > 0 \) and concave upward when \( x < 0 \).
05

Analyze the Behavior

The function \( f(x) = \tan^{-1}(x) \) increases towards its horizontal asymptotes at \( y = \pi/2 \) and \( y = -\pi/2 \). \( f'(x) \) indicates the rate of increase slows as \( x \) moves away from zero. \( f''(x) \) changes sign at \( x = 0 \), confirming the concavity behavior observed from the graph.

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Key Concepts

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

Arctan Function
The arctan function, denoted as \(f(x) = \tan^{-1}(x)\), plays a crucial role in trigonometry and calculus. This function is the inverse of the tangent function and is used to determine angles within the range \((-\pi/2, \pi/2)\). When you input a value into \(\tan^{-1}(x)\), it gives the angle whose tangent is \(x\).
  • For example, \(\tan^{-1}(1) = \pi/4\) because the tangent of \(\pi/4\) is 1.
  • Understanding the range of arctan is vital because it indicates the limits of the output values.
Graphically, \(\tan^{-1}(x)\) represents a curve that asymptotically approaches the lines \(y = \pi/2\) and \(y = -\pi/2\), meaning it gets infinitely close to these lines but never actually touches them. This makes \(\tan^{-1}(x)\) function particularly useful in scenarios where we need calculations within confined ranges, such as angles in navigation and physics.
Derivatives
Calculating the derivatives of a function is like uncovering its underlying mechanics. For the function \(f(x) = \tan^{-1}(x)\), the first derivative \(f'(x)\) provides information on how \(f(x)\) changes:
  • \(f'(x) = \frac{1}{1+x^2}\)
This result tells us the slope at any point on the \(\tan^{-1}(x)\) curve. A positive slope indicates that the function is increasing, which is consistent across all \(x\) values for \(\tan^{-1}(x)\).

The second derivative \(f''(x)\), defined as:
  • \(f''(x) = \frac{-2x}{(1+x^2)^2}\)
helps discern changes in the slope, a concept known as concavity. In examining these derivatives, you can predict how \(f(x)\) behaves without plotting numerous points.
Concavity
Concavity refers to the way a curve bends and the direction in which it "opens". For \(f(x) = \tan^{-1}(x)\), the second derivative \(f''(x)\) offers insights into this property.
  • \(f''(x)<0\) when \(x>0\), indicating that the function is concave downward.
  • Conversely, \(f''(x)>0\) when \(x<0\), showing that the function is concave upward.
This switch in concavity at \(x=0\) highlights a critical point on the graph where the function changes how it bends. Understanding concavity helps in visualizing how the function approaches its asymptotes. When plotting \(\tan^{-1}(x)\), these characteristics give you a detailed preview of whether it will "cup" upwards or downwards, enhancing the precision of sketches and predictions.
Graphing Utility
Graphing utilities are crucial tools in visualizing complex mathematical functions like \(\tan^{-1}(x)\). These utilities take arcane mathematical concepts and present them in clear, visual formats that are easier to understand.
  • Using a graphing utility, you can plot \(f(x) = \tan^{-1}(x)\) along with its first and second derivatives.
  • This graphical representation shows the increasing nature of \(\tan^{-1}(x)\) and the asymptotic behavior leading to its horizontal asymptotes.
Visualizing \(f'(x)\) on the graph reveals how the slope changes, while \(f''(x)\) helps you see where the curve changes direction due to concavity shifts. With graphing utilities, seeing becomes believing, and you can literally see the math in action, making complex theoretical concepts much more tangible and accessible.

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Most popular questions from this chapter

Show that \(e^{x}\) grows faster as \(x \rightarrow \infty\) than any polynomial $$a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$$.

L'H贸pital's Rule does not help with the limits. Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow 0^{+}} \frac{x}{e^{-1 / x}}$$

When is a polynomial \(f(x)\) of smaller order than a polynomial \(g(x)\) as \(x \rightarrow \infty ?\) Give reasons for your answer.

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=(\sqrt{t})^{t}$$

Which one is correct, and which one is wrong? Give reasons for your answers. $$\begin{aligned} a. \lim _{x \rightarrow 0} \frac{x^{2}-2 x}{x^{2}-\sin x} &=\lim _{x \rightarrow 0} \frac{2 x-2}{2 x-\cos x} \\ &=\lim _{x \rightarrow 0} \frac{2}{2+\sin x}=\frac{2}{2+0}=1 \end{aligned}$$ $$\text { b. } \lim _{x \rightarrow 0} \frac{x^{2}-2 x}{x^{2}-\sin x}=\lim _{x \rightarrow 0} \frac{2 x-2}{2 x-\cos x}=\frac{-2}{0-1}=2$$
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