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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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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. In guideline D find an equation of the slant asymptote.

y = 1 − x + e1+x/3

To determine

To sketch: The graph of the function y=1x+e1+x3 and to find the equation of the slant asymptote.

Explanation

Given:

The equation of the curve is given below.

f(x)=1x+e1+x3 (1)

Calculation:

Compute the domain.

Since f(x) is a polynomial function which is defined for all values of x, and so the domain is D=.

Therefore, domain for the variable x is in the limit (,).

Compute the value of y intercept.

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

f(0)=10+e1+03f(0)=3.718

Compute the value of x intercept.

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

f(x)=01x+e1+x3=01x+e1+x30[ex>0]

So, the curve has no x intercept.

Therefore, the given curve has no x intercept and y intercept is 2.

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

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

f(1)=1(1)+e1+(1)3f(1)=3.95

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

f(1)=1(1)+e1+(1)3f(1)=3.79

The conditions f(x)f(x) and f(x)f(x) are true.

Therefore, the given curve is neither symmetric about y-axis nor about the origin.

Compute the vertical asymptote:

The domain is the set of all real numbers and so there is no vertical and horizontal asymptote.

Compute the slant asymptote as follows.

Rewrite equation (1).

f(x)=1x+e1+x3f(x)(1x)=e1+x3

f(x)(1x)=0[limxe1+x3=0]

Therefore, the slant asymptote is y=(1x).

Compute the interval of increase or decrease.

Differentiate equation (2) as follows.

f'(x)=ddx(1x+e1+x3)f'(x)=01+e1+x3ddx[x3+1]f'(x)=1(13ddx[x]+ddx[1])e1+x3f'(x)=1+e1+x33

Check if the condition f'(x)>0 is true and the curve is increasing.

Substitute x=2(ln31) in the equation f'(x)=1+e1+x33.

f'(2(ln31))=1+e1+2(ln31)33f'(2(ln31))=0.438

f'(2(ln31))=0.438>0

Hence, the condition is true and the curve is increasing.

The function f'(x) is increasing in the interval (3(ln31),).

Check if the condition f'(x)<0 is true and curve is decreasing.

Substitute x=3 in the equation f'(x)=1+e1+x33.

f'(3)=1+e1+333f'(3)=0

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