   Chapter 3.5, Problem 53E

Chapter
Section
Textbook Problem

# 49–54 Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asymptote. y = 2 x 3 + x 2 + 1 x 2 + 1

To determine

To sketch:

The curve of y and use guidelines D to find an equation of the slant asymptote

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) Given:

fx=y=2x3+x2+1x2+1

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

For all real values, the function is defined. Therefore, the domain is all real numbers.

Domain R

B. Intercepts

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

y0=203+02+102+1=1

y-intercept is (0, 1)

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

0=2x3+x2+1x2+1

Solve for x

2x3+x2+1=0

x=-1

x intercept is -1, 0

C. Symmetry

For symmetry, replace each x by -x therefore,

f-x=2-x3+-x2+1-x2+1

f-x=-2x3+x2+1x2+1

That means there is no symmetry

D. Asymptote

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

limn±2x3+x2+1x2+1 =±

So, there are no horizontal asymptotes.

Vertical asymptotes: the points where f is not defined.

So there is no vertical asymptote.

To find slant asymptote-

Use long division method rule,

So,fx=2x+1+-2xx2+1

By guidelines of slant asymptote, slant asymptote is y=2x+1

E. Intervals of increase or decrease

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

f'x=2x2(x2+3)x2+12

Equating this derivative with 02x2(x2+3)x2+12=0

Solve for x

x=0

Combining the critical points with the domain, the function has two intervals,

- , 0, (0, )

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

For -, 0 consider x=- 1

f'-1=2-12(-12+3)-12+12=2

f'-1>0

-<x<0, f'x>0

The function is increasing in the interval (-, 0)

For 0,    take x=1

f'1=212(12+3)12+12=2

0<x<, f'x>0

The function is increasing in the interval 0,

Hence, fx is increasing on (-, 0) and ( 0,)

Since f continues on R, f(x) is increasing on -,

F

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