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Chapter 3: Problem 51

Find all points of inflection, if they exist. $$ f(x)=\tan x+\sec x $$

Short Answer

Expert verified
Possible inflection points given by \( x = \arctan(-\frac{1}{3}) + n\pi \) for integers \(n\).

Step by step solution

01

Understand Points of Inflection

Points of inflection occur where the second derivative of the function changes sign. To find them, we need the second derivative and identify where it transitions from positive to negative or vice versa.
02

Find the First Derivative

First, we find the first derivative of the function. Given the function: \[ f(x) = \tan x + \sec x \] The first derivative is: \[ f'(x) = \sec^2 x + \sec x \tan x \]
03

Find the Second Derivative

Next, we find the second derivative of the function: \[ f''(x) = \frac{d}{dx}(\sec^2 x + \sec x \tan x) = \frac{d}{dx}(\sec^2 x) + \frac{d}{dx}(\sec x \tan x) \] Using the chain and product rules: \[ = 2 \sec^2 x \tan x + \sec x(\sec^2 x + \sec x \tan x)) = 2 \sec^2 x \tan x + \sec^3 x + \sec^2 x \tan x = 3 \sec^2 x \tan x + \sec^3 x \]
04

Find Potential Inflection Points

Set the critical points by solving: \[ 3 \sec^2 x \tan x + \sec^3 x = 0 \] Factor out \(\sec^2 x\) from the equation: \[ \sec^2 x (3 \tan x + \sec x) = 0 \] Solve both factors: \[ \sec^2 x = 0 \] (no solution as \(\sec xeq 0\)) \[ 3 \tan x + \sec x = 0 \rightarrow \sec x = - 3 \tan x \rightarrow \frac{1}{\cos x} = -3 \frac{\sin x}{\cos x} \rightarrow \cos x = -3 \sin x \rightarrow \tan x = -\frac{1}{3} \]
05

Verify the Points of Inflection

Verify if the second derivative changes sign at \( \tan x = -\frac{1}{3} \). Given \( x = \arctan(-\frac{1}{3}) + n\pi \) where \(n \in \mathbb{Z} \). Check the behavior of \( f''(x) \) on intervals around these points. If it changes sign, it's a point of inflection.

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Key Concepts

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

Second Derivative
The second derivative, denoted as \(f''(x)\), is a critical tool in calculus used to analyze the concavity of a function. For the function \(f(x) = \tan x + \sec x\), we first find its second derivative to detect points of inflection. This involves taking two derivatives successively. The first derivative, \(f'(x)\), represents the rate of change of \(f(x)\). Calculating it, we get \(f'(x) = \sec^2 x + \sec x \tan x\). The second derivative requires applying differentiation rules again: using the chain rule and product rule, we get \(f''(x) = 3 \sec^2 x \tan x + \sec^3 x\). By solving \(f''(x) = 0\) and verifying sign changes around these points, we can determine the existence of inflection points.
Trigonometric Functions
Trigonometric functions like \(\tan x\) and \(\sec x\) are fundamental in calculus for modeling periodic behaviors. In our problem, the function \(\tan x + \sec x\) uses these functions, which have properties that influence the derivative calculations. The derivative of \(\tan x\) is \(\sec^2x\), and the derivative of \(\sec x\) is \(\sec x \tan x\). Knowing these derivatives helps us efficiently compute higher-order derivatives and analyze the function's behavior over its domain.
Calculus Analysis
Calculus analysis involves studying derivatives and their implications for a function. In this task, we used the first and second derivatives of \(\tan x + \sec x\) to find points of inflection. Calculus helps identify where a function changes from concave up to concave down or vice versa. This change is critical for understanding the function's graph and underlying behavior. By studying where \(f''(x)\) changes sign, we pinpoint inflection points, offering insights into the function's curvature and trends.
Chain Rule
The chain rule is a method in calculus for finding the derivative of composite functions. When dealing with \(\tan x + \sec x\), especially in calculating the second derivative, the chain rule is indispensable. For example, \(\frac{d}{dx}(\sec^2x)\) requires the chain rule: \(2 \sec^2 x \tan x\). This rule helps in breaking down complex derivatives into manageable steps, ensuring accurate results.
Product Rule
The product rule is used when differentiating products of functions. It's essential for our function involving \(\tan x +\sec x\). While calculating \(\frac{d}{dx}(\sec x \tan x)\), the product of \(\sec x\) and \(\tan x\) requires the product rule: \(\sec x(\tan x) + \sec x(\sec x + \tan x)\). This ensures each component is differentiated correctly, and the overall derivative is accurate. Using both chain and product rules, we derive \(f''(x) = 3 \sec^2 x \tan x + \sec^3 x\), leading to our inflection point analysis.

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Most popular questions from this chapter

A ladder 26 ft long leans against a vertical wall. If the lower end is being moved away from the wall at the rate of \(5 \mathrm{ft} / \mathrm{sec}\), how fast is the height of the top decreasing (this will be a negative rate) when the lower end is \(10 \mathrm{ft}\) from the wall?

A pitcher's earnedrun average (the average number of runs given up every 9 innings, or 1 game) is given by $$ E=9 \cdot \frac{n}{i}, $$ where \(n\) is the number of earned runs allowed and is the number of innings pitched. Suppose that we fix the number of earned runs allowed at 4 and let \(i\) vary. We get a function given by $$ E(i)=9 \cdot \frac{4}{i}. $$ a) Complete the following table, rounding to two decimal places. $$ \begin{array}{|l|l|} \hline \begin{array}{l} \text { Innings } \\ \text { Pitched (i) } \end{array} & \begin{array}{l} \text { Earncd-Run } \\ \text { Average (E) } \end{array} \\ \hline 9 & \\ \hline 8 & \\ \hline 7 & \\ \hline 6 & \\ \hline 5 & \\ \hline 4 & \\ \hline 3 & \\ \hline 2 & \\ \hline 1 & \\ \hline \frac{2}{3} \text { (2 outs) } & \\ \hline \frac{1}{3} \text { (l out) } & \\ \hline \end{array} $$ b) Find \(\lim _{i \rightarrow 0^{+}} E(i)\). c) On the basis of parts (a) and (b), determine a pitcher's earned-run average if 4 runs were allowed and there were 0 outs.

For a certain speaker of the vowel in the word hot, \(A_{m}=7 A_{t}, L_{m}=8 \mathrm{~cm}\), and \(L_{t}=9 \mathrm{~cm} .\) Using the two-tube model, find the first three natural frequencies of the vocal tract.

Minimize \(Q=3 x+y^{3}\), where \(x^{2}+y^{2}=2\)

Use a grapher to graph each of the following equations. On most graphers, equations must be solved for \(y\) before they can be entered. \(x^{2}+y^{2}=4\) Note: You will probably need to sketch the graph in two parts: $$ y=\sqrt{4-x^{2}} \text { and } y=-\sqrt{4-x^{2}}. $$ Then graph the tangent line to the graph at the point \((-1, \sqrt{3})\).
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