BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.3, Problem 31E
To determine

To find: The values of x in which the graph of f(x) have a horizontal tangent.

Expert Solution

Answer to Problem 31E

The values of x in which the graph of f(x) have a horizontal tangent is x=2π3+2πn (or) x=4π3+2πn, where n is an integer.

Explanation of Solution

Given:

The function is f(x)=x+2sinx.

Derivative rules:

(1) Sum Rule: ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]

(2) Constant Multiple Rule: ddx[cf(x)]=cddx[f(x)]

Calculation:

Obtain the derivative of f(x).

Apply the sum rule (1),

ddx[x+2sinx]=ddx[x]+ddx[2sinx]

Apply the constant multiple rule (2),

 ddx[x+2sinx]=ddx[x]+2ddx[sinx] =[1]+2[cosx]=1+2cosx

Thus, the derivative of  f(x)=x+2sinx is f(x)=1+2cosx.

Note that, the function f(x) has horizontal tangent when f(x)=0.

1+2cosx=02cosx=1cosx=12x=cos1(12)

Thus, the values of x is  x=2π3+2πn or x=4π3+2πn, n is an integer.

Therefore, the values of x in which the graph of f(x) have a horizontal tangent is x=2π3+2πn (or) x=4π3+2πn, n is an integer.

Note: 4π3 and 2π3 are ±π3 units from π, the value of x is equal to (2n+1)π±π3, where n is an integer.

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