   Chapter 3.5, Problem 18E

Chapter
Section
Textbook Problem

# 1-40 Use the guidelines of this section to sketch the curve. y = x x 3 − 1

To determine

To sketch:

The curve of the given function

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:

y=x x3-1

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 x3-1 is a rational expression, the denominator will be zero at x=1.

Therefore, the domain is -, 1(1, )

B) Intercepts

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

y= 00-1

y=0

y  intercept is 0, 0.

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

0= x x3-1

x=0.

x  Intercept is (0, 0).

C) Symmetry

For symmetry, replace each x by -x.

f-x=-x(-x)3-1=-x-x3-1=xx3+1

f-x- f(x)f(x)

Function is neither even nor odd. Therefore, no symmetry.

D) Asymptote

Horizontal asymptotes

limx-y=limx-x x3-1=0 ,  limx+y=limx+x x3-1=0

Horizontal asymptote is y=0.

Vertical asymptotes

fx=x x3-1

The denominator become zero when x= 1, so it has vertical asymptote at x= 1.

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= -2x3-1x3-12

Equating this derivative with 0,

-2x3-1x3-12=0

Solving this for x, it becomes

x=-12 3,

This is the critical point.

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

The intervals are -,-123,-123,1, 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 - , -123, consider x= -1.

f'-1= -2(-1)3-1(-1)3-12

f'-1=14

-<x<-123; f'x>0

The function is increasing in the interval (-,-123).

For -123 ,1, consider x=0.

f'0= -2(0)3-1(0)3-12

f'0=-1

-123<x<1; f'x>0

The function is increasing in the interval (-123, 1)

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