# All points on the graph such that the tangent line is horizontal. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 3.4, Problem 49E
To determine

## To find: All points on the graph such that the tangent line is horizontal.

Expert Solution

The required points such that the tangent line is horizontal are at (π2+2nπ,3)_ and (2nππ2,1)_, where n is integer.

### Explanation of Solution

Given:

The function is f(x)=2sinx+sin2x .

Derivative Rule: Difference Rule

If f(x).and g(x) are both differentiable function, then

ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)] (1)

Calculation:

Obtain the derivative of f(x).

f(x)=ddx(f(x))=ddx(2sinx+sin2x)

Apply the difference rule as shown in equation (1),

f(x)=ddx[2sinx]+ddx[sin2x]=2ddx[sinx]+ddx[1cos2x2]       ( sin2x=1cos2x2)=2(cosx)+ddx(12)ddx(cos2x2)=2cosx+(0)(sin2x22)

Simplify further and obtain the derivative,

f(x)=2cosx+sin2x=2cosx+2sinxcosx    (sin2x=2sinxcosx )=2cosx(1+sinx)

Thus, the derivative of   f(x)=2sinx+sin2x is f(x)=2cosx(1+sinx) .

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

f(x)=02cosx(1+sinx)=0cosx(1+sinx)=0cosx=0 or 1+sinx=0

Here, it is sufficient to find the value of x for which cosx=0 and it is expressed as follows,

cosx=cos(π2)

Therefore, the general solution for cosx=0 is x=2nπ±π2.

Substitute x=2nπ+π2 in f(x),

f(2nπ+π2)=2sin(2nπ+π2)+sin2(2nπ+π2)=2(sin2nπcosπ2+cos2nπsinπ2)+(sin2nπcosπ2+cos2nπsinπ2)2 =2(0+1)+(0+1)2=3

Thus, the tangent is horizontal at the point is (2nπ+π2,3).

Substitute x=2nππ2 in f(x),

f(2nππ2)=2sin(2nππ2)+sin2(2nππ2)=2(sin2nπcosπ2cos2nπsinπ2)+(sin2nπcosπ2cos2nπsinπ2)2 =2(01)+(01)2=2+1=1

Thus, the tangent is horizontal at the point is (2nππ2,1).

Therefore, the required points such that the tangent line is horizontal are at (π2+2nπ,3)_ and (2nππ2,1)_, where n is integer.

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