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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

Find the relative maxima, relative minima, points of inflection, and asymptotes, if they exist, for each of the functions in Problems 1-3. Graph each function.

y = 4 x 3 x 4 10

To determine

To calculate: The relative minimum, relative maximum, points of inflection and asymptotes for the function y=4x3x410 and sketch its graph.

Explanation

Given Information:

The provided function is,

y=4x3x410

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 is 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 function,

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

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

3. Calculate the sign of the function to the left and right of the critical values.

(a) If f(x)<0 and f(x)>0 on the left and right side respectively of a critical value, then the point is relative minimum.

(b) If f(x)>0 and f(x)<0 on the left and right side respectively of a critical value, then the point is relative maximum.

Calculation:

Consider the provided function,

y=4x3x410

Now, calculate the first derivative.

y=4x3x410y=12x24x3=4x2(3x)

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

4x2(3x)=0

Thus, either (3x)=0 or x2=0.

First consider (3x)=0.

Add x to both sides and simplify:

3x=03x+x=0+x3=xx=3

And,

x2=0x=0

Thus, the critical values of the function are at x=0 and x=3.

Now, substitute 0 for x in the equation y=4x3x410,

y=4(0)3(0)410=10

Substitute 3 for x in the equation y=4x3x410,

y=4(3)3(3)410=1088110=17

Thus, the critical points are (0,10) and (3,17).

Now, calculate the sign of the y' on either side of the critical point.

First consider x=0.

Calculate y at x=1.

y=12x24x3=12(1)24(1)3=12+4=16

Thus, y>0 on left of x=0.

Calculate y at x=1:

y=12x24x3=12(1)24(1)3=124=8

Thus, y>0 on right of x=0

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