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4th Edition

James Stewart

Publisher: Cengage Learning

ISBN: 9781337687805

Chapter 4, Problem 10RE

**(a)**

To determine

**To find: **Thevertical and horizontal asymptote of the function.

Expert Solution

The Horizontal asymptote is

**Given: **

**Concept used: **

If the degree of the numerator is less than the denominator, thehorizontal asymptote is

If the denominator has no zeroes then there has no vertical asymptotes

Or

To get Vertical asymptote function should be rational and denominator must contain some variable otherwise there has no vertical asymptote.

**Calculation:**

Accordingto the laws of asymptotes:

If the degree of the numerator is more than the denominator, there is no horizontal asymptote.

If the denominator has no zeroes then there has no vertical asymptotes/.

Here the numerator of the function has degree 1 and denominator has degree of 2.

the horizontal asymptote is

Vertical asymptote is at

Hence, the Horizontal asymptote is

**(b)**

To determine

**To find: **TheInterval of increasing or decreasing of the function.

Expert Solution

The Interval of increasing or decreasing of the function is

Decreasing at interval of

Increasing at interval of

**Given: **

**Concept used: **

Increasing or decreasing function can be calculated by equating first derivative of the function to 0.

Zeroes of x can be calculatedafter that the increasing and decreasing can be measured.

**Calculation:**

Increasing or decreasing function can be calculated by equating first derivative of the function to 0.

Hence the Interval of increasing or decreasing of the function is

Decreasing at interval of

Increasing at interval of

**(c)**

To determine

**To find: **The Local maxima and minima of the function.

Expert Solution

Local minima.

the point of inflection at

**Given**:

**Concept used: **

The local maxima and minima can be calculated by firstly equating the double differentiation to 0.

1.

2.If

3.

**Calculation:**

At

Hence,

Local minima.

the point of inflection at

**(d)**

To determine

**To find: **The interval of concavity and the inflection point.

Expert Solution

Concave downward in the interval of

Concave upward in the interval of

These points are point of inflection

**Given: **

**Concept used: **

The second derivative of function is calculated first.

Set the second derivative equal to zero and solve.

Check whether the second derivative undefined for any values of x.

Plot the number on number line and test the regions with the second derivative.

Plug these 3 values for obtain three inflection points.

The graph of

The graph of

If the graph of

**Calculation:**

This two are the point of inflection.

By putting the values in the equation.

The interval will be

Hence,

Concave downward in the interval of

Concave upward in the interval of

These points are point of inflection

**(e)**

To determine

**To Sketch:**the graph of the function using graphing device.

Expert Solution

Through the graph it’s easily verified the point of local maxima and minima, function is increasing or decreasing, concavity down or up and point of inflection.

**Given: **

**Concept used: **

Desmos graphing calculator is used her to plot the graph and it can easily verify the maxima, minima and point of inflection etc.

**Calculation:**

The graph of

Hence, through the graph it’s easily verified the point of local maxima and minima, function is increasing or decreasing, concavity down or up and point of inflection.