   Chapter 4.6, Problem 35E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Describe how the graph of f varies as c varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when c changes. You should also identify any transitional values of c at which the basic shape of the curve changes.f(x) = cx + sin x

To determine

To describe: How the graph of f varies as c varies.

Explanation

Given information:

The function is f(x)=cx+sinx .

Calculation:

Show the function as follows:

f(x)=cx+sinx (1)

Differentiate both sides of the Equation (1).

f(x)=c+cosx

Differentiate both sides of the Equation.

f(x)=sinx

Substitute x for x in Equation (1).

f(x)=c(x)+sin(x)=cxsinx=(cx+sinx)=f(x)

Hence, the function is an odd function. So the graph is symmetric about the origin.

Sketch the function for different values of c as shown in Figure 1.

Equating the function is equal to zero

f(x)=0cx+sinx=0sinx=cx

So the value of x-intercept is 0.

Equating the derivative of the function is equal to zero.

f(x)=0c+cosx=0cosx=c

So there is no critical number when |c|>1 .

Refer to Figure 1.

If |c|1 , then there are infinitely many critical numbers.

If x1 is the unique solution of cosx=c in the interval [0,π] , then the critical numbers are 2nπ±x1 , when n ranges over the integers.

Providing f(x)<0 will provides sinx>0 . So the function f is concave downward on intervals of the form (2nπ,(2n+1)π) and concave upward on intervals of the form ((2n1)π,2nπ) .

The inflection points of the function f are the points (nπ,nπc) ,where n is an integer.

If c1 , then f(x)0 for all x, so the function f is increasing and has no extremum.

If c1 , then f(x)0 for all x, so the function f is decreasing and has no extremum

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