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Sketch the graph of f, label the relative extrema, point of inflection and the asymptotes.

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(a) f(x) = (5 − 3x) /( 3x − 2)
 
(b) f(x) = x^2 +(1/x)
 
(c) f(x) = 4x^3 − x^4
 
(d) f(x) = 1/(1 + x^2)
asked Oct 5, 2014 in CALCULUS by anonymous

4 Answers

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a) The rational function

The graph of rational functions can be recognised by the fact two or more parts.

1) Find zeros

y = (5 - 3x) / (3x - 2)

To find  y intercept  x = 0 in the rational function.

y = [5 - 3(0)] / [3(0) - 2]

y = -5/2

y intercept is (0, -5/2)

To find x intercept let the numarator = 0

5 - 3x = 0

- 3x  = -5

x = 5/3

x intercept is (5/3, 0)

2)Find asymptotes

Vertical asymptote can be found by making denominator = 0.

3x - 2 = 0

3x = 2

x = 2/3

Vertical asymptote is x = 2/3.

To find horizontal asymptote, first find the degree of the numarator and  the degree of denominator.

Degree of the numarator =1 and the degree of denominator = 1.

Since the degree of the numerator is equal to the degree of the denominator,horizontal asymptote is the ratio of the leading coefficient of numarator and denominator.

Leading coefficient of numarator = -3, leading coefficient of denominator = 3

Horizontal asymptote is y = -1

We need some more points to more accurate graph.

Choose the random values for x, compute the corresponding y  values.

x

y = (5 - 3x) / (3x - 2)

(x , y)
-2

y = [5 - 3(-2)] / [3(-2) - 2] = 11/-8

(-2, -1.375)
-4

y = [5 - 3(-4)] / [3(-4) - 2] = 17/-14

(-4 , -1.214)
1

y = [5 - 3(1)] / [3(1) - 2] = 2/1

(1, 2)
4

y = [5 - 3(4)] / [3(4) - 2] = -7/10

(1, -0.7)

 

answered Oct 9, 2014 by david Expert
0 votes

Contd...

Graph

1) Draw the coordinate plane.

2) Next dash the horizontal and vertical asymptotes.

3) Plot the x,y intercepts and coordinate pairs found in the table..

4) Connect the plotted points .

When you draw your graph, use smooth curves complete the graph.

To find relative extrema equate f'(x) = 0

To find point of inflection equate f''(x) = 0

image

In this case no relative extrema and inflection points.

 

answered Oct 9, 2014 by david Expert
0 votes

(b)

The rational function f(x) = x2 + (1/x)

y = (x3 + 1)/x

The graph of rational functions can be recognised by the fact two or more parts.

1) Find zeros

y = (x3 + 1)/x

To find  y intercept  x = 0 in the rational function.

y = (03 + 1)/0

In this case there is no y intercept .

To find x intercept  y = 0 in the rational function.

0 = (x3 + 1)/x

x3 + 1 = 0

x3 = - 1

x = - 1

x intercept is (-1, 0)

2)Find asymptotes

Vertical asymptote can be found by making denominator = 0.

x = 0

Vertical asymptote is x = 0.

To find horizontal asymptote, first find the degree of the numarator and  the degree of denominator.

Degree of the numarator = 3 and the degree of denominator = 1.

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

We need some more points to more accurate graph.

Choose the random values for x, compute the corresponding y values.

x

y = (x3 + 1)/x

(x , y)
-2

y = [(-2)3 + 1]/(-2) = -7/-2

(-2, 3.5)
-0.5

y = [(-0.5)3 + 1]/(-0.5) = -1.75

(-0.5 , -1.75)
1

y = [(1)3 + 1]/1 = 2

(1, 2)
2

y = [(2)3 + 1]/2 = 9/2

(2, 4.5)

 

answered Oct 9, 2014 by david Expert
0 votes

Contd...

Graph

1) Draw the coordinate plane.

2) Next dash the vertical asymptotes.

3) Plot the x intercept and coordinate pairs found in the table..

4) Connect the plotted points .

When you draw your graph, use smooth curves complete the graph.

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

f'(x) = [x(3x2) - ((x3 + 1)(1)]/x2

f'(x) = [3x3 - (x3 + 1)]/x2

f'(x) = [3x3 - x3 - 1]/x2

f'(x) = [2x3 - 1]/x2

f''(x) = [x2(6x2) - (2x3 - 1)(x2)]/x4

f''(x) = [6x4 - (2x5 - x2)]/x4

f''(x) = [6x4 - 2x5 + x2)]/x4

f''(x) = x2[6x2 - 2x3 + 1]/x4

f''(x) = [6x2 - 2x3 + 1]/x2

To find relative extrema equate f'(x) = 0

[2x3 - 1]/x2 = 0

2x3 - 1 = 0

2x3 = 1

x3  = 1/2

x = 0.7937

f'(x) = [2x3 - 1]/x2

= [2(0.79)3 - 1]/(0.79)2 = -0.0139/0.6241 = -0.022 < 0

There is no relative extrema.

To find point of inflection equate f''(x) = 0

 [6x2 - 2x3 + 1]/x2 = 0

6x2 - 2x3 + 1 = 0

There is only one real solution at x = 3.06

f''(x) = [6x2 - 2x3 + 1]/x2

= [6(3.06)2 - 2(3.06)3 + 1]/(3.06)2

= (56.181 - 57.305 + 1)/(9.3636)

= -0.013 < 0

There is no inflection points.

 

answered Oct 9, 2014 by david Expert

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