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Single Variable Calculus

8th Edition
James Stewart
ISBN: 9781305266636

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Single Variable Calculus

8th Edition
James Stewart
ISBN: 9781305266636
Textbook Problem

Use the guidelines of this section to sketch the curve.

35. y = x tan x, −π/2 < x < π/2

To determine

To Sketch: The curve by plotting x and y coordinates using guidelines.

Explanation

Given:

The Cartesian equation is as below.

y=f(x)=xtanx (1)

Calculation:

a)

Calculate the domain.

The function is defined for the given interval π2<x<π2 . So the domain is (π2,π2)

b)

Calculate the intercepts.

Calculate the value of x -intercept.

Substitute 0 for y in the equation (1).

xtanx=0x=0

Hence, x -intercept is (0,0) .

Calculate the y -intercept.

Substitute 0 for x in the equation (1).

f(0)=0tan(0)=0

Therefore, the y -intercept is (0,0) .

c)

Calculate the symmetry.

Apply negative and positive values for x in the equation (1).

Substitute 1 for x in the equation (1).

f(1)=(1)tan(1)=1.557

Substitute +1 for x in the equation (1).

f(1)=(1)tan(1)=1.557

Here the condition f(x)=f(x) is true, hence it is an even function and the graph is symmetrical about y -axis.

d)

Calculate asymptotes.

Apply limit of x tends to π2 in the equation (1).

limxπ2y=π2tan(π2)=π2()=

Apply limit of x tends to π2 in the equation (1).

limxπ2y=π2tan(π2)=π2()=

Here, the value of limit gets infinity; this implies there is no horizontal asymptote.

For the given domain (π2,π2) , there will be vertical asymptotes at x=±π2

e)

Calculate the intervals.

Differentiate the equation (1) with respect to x .

Apply the product rule below.

(uv)'=uv'+vu'

Substitute x for f and tanx for v in the above equation.

f'(x)=xsec2x+tanx(1)

f'(x)=xsec2x+tanx (2)

Substitute 0 for f'(x) in the equation (2).

xsec2x+tanx=0tanx=xsec2x

Multiply cosx on both sides of the above equation.

cosxtanx=xsec2xcosx

Substitute (sinxcosx) for tanx and (1cos2x) for sec2x in the above equation.

cosx(sinxcosx)=x(1cos2x)cosxsinx=xcosxsinxcosx=x

Substitute sin2x2 for sinxcosx in the above equation.

sin2x2=xsin2x=2x

The only solution for the above equation is x=0

Take the interval (0,π2) .

Substitute π6 for x in the equation (2).

f'(π6)=π6sec2(π6)+tan(π6)=π6(1cos2(π6))+tan(π6)=π6(1cos2(π6))+13

f'(π6)=π6(1(32)2)+13=π6(43)+13=0.698+0.577=1.28

Here the condition f'(π6)>0 is true, hence the function f is increasing on (0,π2) .

Take the interval (π2,0) .

Substitute π6 for x in the equation (2)

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