/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 a. For what values of \(x\) does... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 66

a. For what values of \(x\) does \(g(x)=x-\sin x\) have a horizontal tangent line? b. For what values of \(x\) does \(g(x)=x-\sin x\) have a slope of \(1 ?\)

Short Answer

Expert verified
Question: For the function \(g(x) = x - \sin x\), determine (a) the values of \(x\) where the tangent is horizontal, and (b) the values of \(x\) where the tangent has a slope of \(1\). Answer: (a) There is a horizontal tangent for any values of \(x\) such that \(x = 2n\pi\) where \(n\) is an integer. (b) There is a tangent line with a slope of \(1\) for any values of \(x\) such that \(x = (2n + 1)\frac{\pi}{2}\) where \(n\) is an integer.

Step by step solution

01

Find the derivative of \(g(x)\)

To analyze the tangent lines of a function, we need to find the first derivative of the given function. In this case, we have \(g(x) = x - \sin x.\) We can differentiate this function term by term: \[\frac{d}{dx} (x) = 1,\] \[\frac{d}{dx} (-\sin x) = -\cos x.\] Thus, the first derivative of \(g(x)\) is: \[g'(x) = 1 - \cos x.\]
02

Determine the values of \(x\) with a horizontal tangent#line_content#To find the values of \(x\) with a horizontal tangent, we need to solve for \(x\) when the derivative is equal to zero: \[g'(x) = 0 \Rightarrow 1 - \cos x = 0.\] Now, we need to solve for \(x\): \[\cos x = 1.\] The cosine function equals \(1\) when \(x = 2n\pi\) for any integer \(n\). Therefore, there is a horizontal tangent at any of these values of \(x\).

Step 3: Determine the values of \(x\) with a slope of \(1\)
03

To find the values of \(x\) with a slope of \(1\), we need to solve for \(x\) when the derivative is equal to \(1\): \[g'(x) = 1 \Rightarrow 1 - \cos x = 1.\] Now, we need to solve for \(x\): \[\cos x = 0.\] The cosine function equals \(0\) when \(x = (2n + 1)\frac{\pi}{2}\) for any integer \(n\). Therefore, the tangent line has a slope of \(1\) at any of these values of \(x\).

Final Answer#line_content#(a) There is a horizontal tangent for any values of \(x\) such that \(x = 2n\pi\) where \(n\) is an integer. (b) There is a tangent line with a slope of \(1\) for any values of \(x\) such that \(x = (2n + 1)\frac{\pi}{2}\) where \(n\) is an integer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Derivatives
Understanding derivatives is essential in calculus because they measure how a function changes as its input changes. A derivative represents the rate of change of a function with respect to a variable. For example, if you have a function that represents the position of a car over time, the derivative of that function will give you the car's speed, which is the rate of change of its position.

In more mathematical terms, if you have a function such as the distance traveled by a car over time, labeled as \( s(t) \), the derivative of \( s \) with respect to time, written as \( s'(t) \) or \( \frac{d}{dt}s(t) \), will give you the velocity of the car at each moment of time. Derivatives are found using various rules and formulas depending on the complexity of the function. In our exercise, the derivative of \( g(x) = x - \sin x \) is calculated simply by differentiating each term separately.
Slope of a Tangent Line
The slope of a tangent line to a curve at any given point is the value of the derivative of the function at that point. This slope tells us how steep the line is at that precise part of the curve. A horizontal tangent line, which is of particular interest in our exercise, has a slope of zero. This means that at the point where the tangent is horizontal, the function's rate of change with respect to \(x\) is zero.

In practical terms, if the slope is positive, the function is increasing at that point, and if the slope is negative, the function is decreasing. When the slope is zero, as it is for a horizontal tangent line, the function is neither increasing nor decreasing鈥攊t is momentarily constant. In the exercise, to find where \( g(x)\) has a horizontal tangent line, we set the derivative \( g'(x)\) to zero and solve for \(x\).
Trigonometric Functions Differentiation
Differentiating trigonometric functions is a critical operation in calculus, often required to solve a variety of problems involving rates of change in applied contexts, such as physics and engineering. The derivatives of the six basic trigonometric functions (sine, cosine, tangent, cotangent, secant, and cosecant) are derived using the limits of ratios of differences and are part of the foundation of trigonometric calculus.

For instance, the derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \( -\sin x \). These derivatives are applied when finding the slope of the tangent line that touches the curve of a trigonometric function. In the given exercise, the differentiation of \( \sin x \) plays a crucial role in finding both the horizontal tangents and the points where the slope of the tangent is 1. It is through the understanding of these derivatives that we arrived at the final answers.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a. Differentiate both sides of the identity \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\) to prove that \(\sin 2 t=2 \sin t \cos t\). b. Verify that you obtain the same identity for sin \(2 t\) as in part (a) if you differentiate the identity \(\cos 2 t=2 \cos ^{2} t-1\). c. Differentiate both sides of the identity \(\sin 2 t=2 \sin t \cos t\) to prove that \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\).

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x},\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this assumption implies that the rule is true for \(k=n+1\). (Hint: Write \(e^{(n+1) x}\) as the product of two functions and use the Product Rule.)

Find the following higher-order derivatives. $$\frac{d^{n}}{d x^{n}}\left(2^{x}\right)$$

Economists use production functions to describe how the output of a system varies with respect to another variable such as labor or capital. For example, the production function \(P(L)=200 L+10 L^{2}-L^{3}\) gives the output of a system as a function of the number of laborers \(L\). The average product \(A(L)\) is the average output per laborer when \(L\) laborers are working; that is \(A(L)=P(L) / L\). The marginal product \(M(L)\) is the approximate change in output when one additional laborer is added to \(L\) laborers; that is, \(M(L)=\frac{d P}{d L}\). a. For the given production function, compute and graph \(P, A,\) and \(M\). b. Suppose the peak of the average product curve occurs at \(L=L_{0},\) so that \(A^{\prime}\left(L_{0}\right)=0 .\) Show that for a general production function, \(M\left(L_{0}\right)=A\left(L_{0}\right)\).
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.