   Chapter 3.5, Problem 59E

Chapter
Section
Textbook Problem

# Discuss the asymptotic behavior of f ( x ) = x 4 + 1 x in the same manner as in Exercise 58. Then use your results to help sketch the graph of f .

To determine

To discuss:

The asymptotic behavior of fx=x4+1x , and use this fact to help sketch the graph of f

Explanation

1) Concept:

i) A domain is the set of x values that satisfy the function.

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 behaviour of function. Thus, if f-x=fx, then it is an even function, so it has y-axis symmetry. If f-x=-fx, then it is an odd function, so it has x-axis symmetry. 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 if f''x=0, give the values of inflection points

2) Definition:

Slant asymptote: the line y=mx+b is called a slant asymptote if

limx±fx-(mx+b)=0

3) Given:

fx=x4+1x

4) Calculation:

To show limxfx-x3=0

Put fx=x4+1x in the above equation

limxfx-x3

=limxx4+1x-x3

=limx1x

=0

So,y=x3 is a slant asymptote

We 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

Domain is the set of all x values that satisfies the function.

The function is not defined at 0

Therefore Domain= (-,0)(0,)

B. Intercepts

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

y0=04+10=

Therefore, no y-intercept

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

0=x4+1x

Solve x

x=-14

Domain is (-,0)(0,)

Therefore, no x intercept

C. Symmetry

For symmetry, replace each x by -x therefore,

f-x=-x4+1-x

f-x=x4+1-x

fx=-f(x)

That means there is symmetry about origin

D. Asymptote:

Horizontal asymptotes-the degree of numerator is greater than the degree of denominator, so there is no horizontal asymptote.

limn±x4+1x =±

So, there are no horizontal asymptotes.

Vertical asymptotes- the points where f is not defined.

So vertical asymptote is x=0

E. Intervals of increase or decrease

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

f'x=3x2-1x2

Equating this derivative with 03x2-1x2 =0

Solve for x

x=±134

Combining the critical points with the domain, therefore function has four intervals,

- , -134, -134, 0, 0, 134, 134,

Now, take a test point from each of the above intervals, and check whether the function is increasing or decreasing in that interval.

For -, -134 consider x=-2

f'-2=3-22-1-22 =11.75

f'-2>0

-<x<-134 , f'x>0

The function is increasing in the interval -, -134

For -134, 0 consider x=-0

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