   Chapter 3.4, Problem 48E

Chapter
Section
Textbook Problem

# 48-51 Find the horizontal asymptotes of the curve and use them, together with concavity and intervals of increase and decrease, to sketch the curve. y = 1 + 2 x 2 1 + x 2

To determine

To find:

i. The horizontal asymptotes of the curve

ii. Sketch the curve using horizontal asymptote, concavity and intervals of increase and decrease

Explanation

1) Concept:

Use the definition of horizontal asymptote, find intervals of increase and decrease and concavity to sketch the curve

2) Definition:

Horizontal asymptote:

Horizontal asymptote: y=b

As x±,yb or limx±f(x)=b

3) Given:

y=1+2x21+x2

4) Calculation:

i. Consider the given function,

y=1+2x21+x2

Divide numerator and denominator by x2 and by using limit properties find the limit

=limx1+2x21+x2

=limx1+2x2x21+x2x2

Simplify,

=limx1x2+21x2+1

Apply limit separately,

=limx1x2+limx2limx1x2+limx1

Since, 1x20 as x

Therefore,

=0+20+1

=21

=2

In computing the limit as x- for x<0 we have

|1+2x21+x2|=1+2x21+x2

Therefore, from above

limx-1+2x21+x2=2

Therefore, the line y=2 is horizontal asymptote

The vertical asymptote is likely to occur when denominator 1+x2=0

But the denominator is always positive. That is there is no such x for which denominator is 0

Thus, it has no vertical asymptote.

ii

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