   Chapter 3.5, Problem 20E

Chapter
Section
Textbook Problem

# 1-40 Use the guidelines of this section to sketch the curve. y = x 3 x − 2

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 particular interval. And if f''x=0, give the values of inflection points.

2) Given:

y=x3x-2

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=x3x-2   is a rational expression, its domain is -,2(2,).

Because, at x=2, the denominator becomes 0, it makes the function undefined.

B) Intercepts:

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

y= 00-2

y=0

y  intercept is (0,0).

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

0= x3x-2

x=0.

x  intercept is (0, 0).

C) Symmetry

For axis of symmetry, replace x by (-x).

f-x=(-x)3-x-2=-x3-x-2

f-x- f(x)f(x)

The function is neither even nor odd. Hence, the function has no symmetry.

D) Asymptote

Horizontal asymptotes:

limx-y=limx-x3x-2= ,  limxy=limxx3x-2=

So no horizontal asymptote

Vertical asymptotes

fx=x3x-2

Since the function becomes undefined at x=2, x=2 is vertical asymptote.

Slant asymptotes:

Here the degree of the numerator is not exactly 1 greater than the denominator. Therefore, it won’t have 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= 2x3-6x2x-22

Equating this derivative with 0,

2x3-6x2x-22=0

Solving this for x, it gives x=3 and x=0.

By using this critical point and the domain, create intervals as,-,0,0, 2,2, 3, (3, ).

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= 2-13-6-12-1-22

f'-1=-89

-<x<0; f'x<0

The function is decreasing in the interval (-,0).

For 0, 2 consider x=1

f'1= 213-6121-22

f'1=-4

0<x<2; f'x<0

The function is decreasing in the interval (0, 2)

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