   Chapter 3.5, Problem 19E

Chapter
Section
Textbook Problem

# 1-40 Use the guidelines of this section to sketch the curve. y = x 3 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=x3 x3+1

3) Calculation:

Here, first find the domain of the given function, the x & y intercepts, check the symmetry, asymptotes, intervals of increase and decrease, local maximum and minimum values, concavity and points of inflection and then using this sketch the curve.

A) Domain:

Since y=x3 x3+1  is a rational expression, the denominator will be zero when 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= x3 x3+1

x=0

x  Intercept is 0, 0.

C) Symmetry

For symmetry, replace each x by -x.

f-x=-x3(-x)3+1=-x3-x3+1=x3x3-1

f-x- f(x)f(x)

So, the function is neither even nor odd. Therefore, no symmetry.

D) Asymptote

Horizontal asymptotes:

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

Horizontal asymptote is y=1.

Vertical asymptotes:

fx=x3 x3+1

The domain 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, we first find the derivative of the given function.

f'x= 3x2 (x3+1)2

Equating this derivative with 0,

3x2 (x3+1)2=0

Solving this for x, so it becomes x=0.

This is a critical point.

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

The intervals are -,-1,-1,0, (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 - , -1, consider x= -2.

f'-2= 3(-2)2 ((-2)3+1)2

f'-2=1249

-<x<-1; f'x>0

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

For -1 ,0, consider x=-12.

f'-12=3-122 (-123+1)2

f'-12=4849

-1<x<0; f'x>0

The function is increasing in the interval (-1, 0)

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