   Chapter 4.5, Problem 42E ### 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 = (1 − x)ex

To determine

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

Explanation

Given:

The given curve is as below.

f(x)=y=(1x)ex (1)

Calculation:

(a)

Calculate the domain.

From the equation (1), the function is defined for all real values of x (there is no restriction on the value of x). Hence, the domain of f(x) is all real values of x .

For the polynomial the domain is (,) .

(b)

Calculate the intercepts.

Calculate the y intercept.

Substitute 0 for the value of x in equation (1).

f(x)=y=(1x)ex

f(0)=(10)e0=1

The y intercept is y=1 .

Calculate the value of x intercept as below.

Substitute 0 for y in the equation (1).

0=(1x)ex(1x)=0x=1

The x intercept is x=1 .

(c)

Calculate the symmetry.

Substitute x for x in the equation (1).

f(x)=(1(x))ex=(1+x)exf(x)f(x)

The function is neither even nor odd. Therefore, the curve has no rotational symmetry about the origin.

(d)

Calculate asymptotes.

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

limxf(x)=(1x)ex=(1)e=

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

limxf(x)=(1x)ex=(1())e=(1())0=0

Therefore, the asymptote is horizontal and it is going to the left y=0 .

(e)

Calculate the intervals.

f(x)=(1x)ex

Differentiate the equation (1) with respect to x .

f'(x)=(1x)ex+ex(1)=exxexexf'(x)=xex

Equate the expression f'(x) to zero.

xex=0x=0

Consider the limit for variable x with range (,0) .

Substitute the value of -1 for x in f'(x) .

f'(1)=(1)e1=1e=0.367>0

Therefore, f shows an increasing behavior for range (,0)

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 