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 3.4, Problem 49E
To determine

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

Expert Solution

Answer to Problem 49E

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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