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Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

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BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

In Problems 19-24, a function and its first and second derivatives are given. Use these to find any horizontal and vertical asymptotes, critical points, relative maxima, relative minima, and points of inflection. Then sketch the graph of each function.

y = 3 x 3  + 1 x

y ' = x 4 / 3 1 x 2

y " = 6 2 x 4 / 3 3 x 3

To determine

To calculate: The horizontal and vertical asymptote, relative minimum, relative maximum and points of inflection for the provided equation y=3x3+1x and the first derivative y'=x431x2 and second derivative y"=62x433x3 and sketch its graph.

Explanation

Given Information:

The provided equation is y=3x3+1x and the first derivative y'=x431x2 and second derivative y"=62x433x3.

Formula used:

A vertical asymptote of a function f(x) is a line x=a such that f(a)=.

A vertical asymptote of a rational function h(x)=f(x)g(x) is x=a where g(a)=0 and f(a)0.

A horizontal asymptote of a function f(x) is a line y=b such that limxf(x)=b or limxf(x)=b.

A horizontal asymptote of a rational function h(x)=f(x)g(x) is

1. A line y=0 if the degree of the numerator is less than the degree of the denominator.

2. The line y= ratio of the leading coefficients if the degree of the numerator is equal to the degree of the denominator.

3. Does not exist if the degree of the numerator is greater than the degree of the denominator.

To find relative maxima and minima of a equation,

1. Set the first derivative of the equation to zero, f'(x)=0, to find the critical values of the equation.

2. Substitute the critical values into f(x) and calculate the critical points.

3. Evaluate f(x) at each critical value for which f(x)=0.

(a) If f(x0)<0, a relative maximum occurs at x0.

(b) If f(x0)>0, a relative minimum occurs at x0.

(c) If f(x0)=0 or f(x0) is undefined, the second derivative test fails and then use the first derivative test.

Calculation:

Consider the provided function,

y=3x3+1x

Recall that a vertical asymptote of a rational function h(x)=f(x)g(x) is x=a where g(a)=0 and f(a)0.

Set the denominator of the function equal to zero.

x=0

Thus, vertical asymptote is x=0.

Consider limxf(x).

limxf(x)=limx(3x3+1x)=

Thus, there is no horizontal asymptote.

Now, consider the first derivative of the function y'=x431x2.

Now, to obtain the critical values, set y'=0 as,

x431x2=0x431=0x43=1x=±1

Thus, x=1 or x=1.

Thus, the critical points are x=1 and x=1.

Substitute 1 for x in the equation y=3x3+1x,

y=3x3+1x=313+11=3+1=4

Thus, the critical point is (1,4).

Substitute 1 for x in the equation y=3x3+1x,

y=3x3+1x=313+11=31=4

Thus, the critical point is (1,4)

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