   Chapter 3.5, Problem 35E

Chapter
Section
Textbook Problem

# 1-40 Use the guidelines of this section to sketch the curve. y = x tan x ,     − π / 2 < x < π / 2

To determine

To sketch:

The curve of y

Explanation

1) Concept:

i) The domain is the set of x values for which the function is defined.

ii) To find x-intercept, put y=0, and to find y-intercept, put x=0 in the given function.

iii) Symmetry: To find symmetry, replace x by –x and check the behavior of function. Thus, if f-x=fx, then it is an even function, so its graph is symmetric about the y-axis. If f-x=-fx, then it is an odd function, so its graph is symmetric about the origin. And if f-x-fxfx, then it has no symmetry.

iv) An asymptote is a tangent at infinity. To find horizontal, vertical, and slant asymptote, follow the rules.

v) A function is increasing if f'x>0  and decreasing if f'x<0 in that particular interval.

vi) The number f(c) is a local maximum value of f  if fcf(x) when x is near c and is a local minimum value of f if fc f(x) when x is near c.

vii) If f''x>0, the function is concave up and if f''x<0, the function is concave down in that particular interval. And f''x=0, gives the values of inflection points.

2) Given:

y= xtanx, -π2<x<π2

3) Calculations:

Here, first find the domain of the given function and the x & y intercepts. Next, check the symmetry, asymptotes, intervals of increase and decrease, local maximum and minimum values, concavity, and points of inflection. Using these, sketch the curve.

A. Domain

fx is defined for all values of the given interval.

So, domain is -π2<x<π2

B. Intercepts:

For y intercept, plug x=0  in the given function, and solve it for y.

y=0·tan0

The y-intercept is 0, 0.

For x intercept, plug y=0 in the original function, and solve it for x.

0= x·tanx

The x intercept is 0 , 0.

C. Symmetry:

For f-x, replace x by -x.

f-x= -xtan(-x)

f-x= -x(-tanx)

f-x= x(tanx)

f-x= f(x)

f(x) is an even function.

So, it has y-axis symmetry.

D. Asymptote:

a) Horizontal asymptotes

The domain is bounded so, there is no horizontal asymptote.

Now

limn-π2+xtanx=-π2(-)

limn-π2+xtanx=

limnπ2-xtanx=π2()

limnπ2-xtanx=

Therefore,

b) Vertical asymptotes

limn-π2+xtanx=

limnπ2-xtanx=

So, there are vertical asymptotes at x=±π2.

E. Intervals of increase or decrease:

To find the intervals of increase or decrease, find the derivative of the given function.

f'x= xtanx'+x'tanx

f'x= x sec2x+1tanx

f'x= x sec2x+tanx

To find f'=0 or undefined,

0= xsec2x+tanx

xsec2x=-tan x

Multiply both sides by cosx.

xcosx=-sinx

x=-sinxcosx

-2x=2sinxcosx

-2x=sin2x

This is true for only x=0

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