   Chapter 3.5, Problem 16E

Chapter
Section
Textbook Problem

# 1-40 Use the guidelines of this section to sketch the curve. y = ( x − 1 ) 2 x 2 + 1

To determine

To sketch:

The curve of the given function

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

2) Given:

y= x-12 x2+1

3) Calculation:

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

Since y= x-12 x2+1  is a rational expression, and the denominator is never zero for all real x values therefore, the domain is -,.

B) Intercepts

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

y= 0-120+1

y=1

y  intercept: (0, 1)

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

0=  x-12 x2+1

x=1.

x  intercept: (1, 0)

C) Symmetry

For symmetry, replace each x by (-x).

f-x=-x-12(-x)2+1=(x+1)2x2+1

f-x- f(x)f(x)

The function is neither even nor odd; therefore, no symmetry.

D) Asymptote

Horizontal asymptotes

limx-y=limx-(x-1)2x2+1=1 ,  limxy=limx(x-1)2x2+1=1

Horizontal asymptote is  y=1.

Vertical asymptotes

fx=(x-1)2x2+1

The domain is the set all real numbers, so it has no vertical asymptote.

Slant asymptotes

No slant asymptote.

E) Intervals of increase or decrease.

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

f'x= 2x2-2x2+12

Equating this derivative with 0.

2x2-2x2+12=0

Solving this for x, that is, x=±1,

These are the critical points.

By using this critical point and the domain, create intervals as

-, -1, -1, 1  & (1,  )

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

For - , -1, consider x= -2.

f'-2= 2(-2)2-2(-2)2+12

f'-2=625

-<x<-1; f'x>0

The function is increasing in the interval (-,-1)

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