   Chapter 4.5, Problem 20E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

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

To determine

To Find: The curve by plotting x and y coordinates using guideline.

Explanation

Given:

The given curve is y=x3x2 (1)

Calculation:

Rewrite the equation (1).

x2+2x+4_x2)x3x32x2_0+2x22x24x_4x4x8_8

The equation is y=x2+2x+4+8x2 .

(a)

Calculate the domain.

Consider the equation x2 which is equal to zero.

x2=0x=2

Therefore, (,2) and (2,) are the domain range of coordinates.

(b)

Calculate the intercepts.

If the value of x=0 , calculate the value of y .

y=x3x2y=0302y=0

Therefore, intercept of points x and y are zero.

(c)

Calculate the symmetry.

Apply negative and positive values for x in the equation (1).

Substitute -1 for x in the equation (1).

f(x)=x3x2f(1)=(1)3(1)2=13

Substitute the value of 1 for x in the equation (1).

f(x)=x3x2f(1)=(1)3(1)2=11=1

Hence, the condition f(x)f(x) is true. Therefore, there is no symmetry.

(d)

Calculate asymptotes.

Here, the degree of denominator is less than the degree of numerator, so it won’t have horizontal asymptotes, but it has different kind of asymptotes because the difference in degree is greater than one.

Apply limit of x tends to -2 (x2) in the equation (1).

limx21(x3x2)=(23)22=80=

Apply limit of x tends to 2 (x2+) in the equation (1).

limx2+1(x3x2)=(23)22=80=

Therefore, the value of limits gets to infinity when x=2 which is in vertical axis.

(e)

Calculate the intervals.

Differentiate the equation (1) by applying U/V method.

f'(x)=vu'uv'v2f'(x)=x3x2=(x2)(3x2)(x3)(1)(x2)2=3x36x2x3(x2)2=2x36x2(x2)2

f'(x)=2x2(x3)(x2)2 (2)

If the value of 3 is applied for x in the above equation, f'(x) will get zero. Therefore, the value of x should be greater than three for f'(x) to be greater than 0. Then, the condition f'(x)<0 is true, the value of x<0 . Therefore, the interval between x is between the following intervals 0<x<2 and 2<x<3 . The coordinates of the ranges for domain is (,0),(0,2),(2,3)and(3,) .

(f)

The minimum value appears when the condition f'(x)<0 is true, within the intervals and a maximum value appears when the condition f'(x)>0 is true within the intervals.

Calculate the local minimum and maximum values.

Substitute -1 for x between the intervals (,0) in the equation (2).

f'(x)=2x2(x3)(x2)2f'(1)=2(1)2(13)(12)2=0

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