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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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Section
BuyFindarrow_forward

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 + ex)−2

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=(1+ex)2 (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(0)=(1+e0)2=22=0.25

The y –intercept is y=0.25 .

Calculate the x –intercept.

Substitute 0 for the value of y in equation (1)

(1+ex)2=0

ex=1

The value of ex cannot be negative for any value of x.

The x –intercept does not exist.

(c)

Calculate the symmetry.

Substitute x for x in the equation (1).

f(x)=(1+ex)2f(x)f(x)f(x)f(x)

The function is neither even nor odd. So, the curve is not symmetric about y axis and origin.

(d)

Calculate asymptotes.

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

limxf(x)=(1+e)2=2=12=0

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

limxf(x)=(1+e)2=(1+0)2=1

It has horizontal asymptotes, which are y=0 and y=1 .

The value of f(x) cannot be infinity for any values of x, and there cannot be any vertical asymptote for f(x) .

(e)

Calculate the intervals.

f(x)=(1+ex)2=1(1+ex)2

Differentiate the equation (1) with respect to x .

f'(x)=(1+ex)2(0)[1(2)(1+ex)(ex)]((1+ex)2)2

f'(x)=2ex(1+ex)3 (2)

The value of f'(x) will be less than 0 for the values of x inside the range (,) .

The function f(x) is decreasing inside the range (,) .

(f)

Calculate the local minima and local maximum.

From the information of part (e), function f(x) is continuously decreasing inside the domain for x x(,) .

There are no local maxima and minima.

(g)

Calculate the concavity and point of inflection coordinates.

Differentiate equation (2) with respect to x.

f''(x)=2ex(1+ex)3f''(x)=(1+ex)3(2ex)[-2ex3(1+ex)2(ex)]((1+ex)3)2

f''(x)=(1+ex)3(2ex)[-2ex3(1+ex)2(ex)]((1+ex)3)2 (3)

Equate equation (3) to zero

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