Curve Sketching Using Calculus: A Comprehensive Guide, Slides of Calculus for Engineers

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Lesson 5

Analysis of Functions II:

Relative Extrema;

Graphing Polynomials

OBJECTIVES:

• to define maximum, minimum, inflection,

stationary and critical points, relative maximum and

relative minimum;

• to determine the critical, maximum and minimum

points of any given curve using the first and

second derivative tests;

• to draw the curve using the first and second

derivative tests; and

• to describe the behavior of any given graph in

terms of concavity and relative extrema

and arelativemaximumat x 1

hasarelativeminimaat x 1 and x 2

x x 4 x 1

x

y

4 3 2

EXAMPLE

:

y

x

The points x

1

, x

2

, x

3

, x

4

, and x

5

are critical points. Of

these, x

1

, x

2

, and x

5

are stationary points.

FIRST DERIVATIVE TEST

Theorem 4. 2. 2 asserts that the relative extrema

must occur at critical points, but it does not say

that a relative extremum occurs at every critical

point.

A function has a relative extremum at those critical

points where f’ changes sign.

SECOND DERIVATIVE TEST

There is another test for relative extrema that is based

on the following geometric observation:

• a function f has a relative maximum at stationary

point if the graph of f is concave down on an open

interval containing that point

• a function f has a relative minimum at stationary point

if the graph of f is concave up on an open interval

containing that point

Note: The second derivative test is applicable only to

stationary points where the 2

nd

derivative exists.

INTERVAL (3x)(x-2) f’(x) CONCLUSION

x<0 (-)(-) + f is increasing on

0<x<2 (+)(-) - f is decreasing on

x>2 (+)(+) + f is increasing on

INTERVAL (6)(x-1) f’’(x) CONCLUSION

x<1 (-) - f is concave down on

x>1 (+) + f is concave up on

The 2

nd

table shows that there is a point of inflection at

x=1,

since f changes from concave up to concave down at that

point.

The point of inflection is (1,-1).

 

2. Analyzeand tracethecurveof f x 3 5.

5 3

 x  x

SOLUTION :

      

    

   

    f '' x 60x -30x 30x 2x 1

x 0; x -1; x 1

15x 0 ; x 1 0 ; x 1 0

when f ' x 0 15x x 1 x 1 0

f ' x 15x -15x 15x x 1 15x x 1 x 1

3 2

2

2

4 2 2 2 2

  

  

    

    

     

y

x





3. Analyze and trace the curve of y  3 x  x

    

   

x -1 and x 1

1 x 0 and 1 x 0

3 1 - x 3 1 x 1 x 0

y' 3 3 x

y 3 x x

2

2

3

 

   

   

 

 

x 0

6 x 0

y'' 6 x



 

 

SOLUTION :



y

x

3

y  3 x  3 x

2

4 x

4 x

3. Analyze and trace the curve of y

    

 

   

 

 

  

 

   

x -2 and x 2

2 x 0 and 2 x 0

0

4 x

4 2 x 2 x

4 x

4 4 x

y'

4 x

16 4 x

4 x

16 4 x 8 x

y'

4 x

4 x 4 4 x 2 x

y'

4 x

4 x

y

2 2 2 2

2

2 2

2

2 2

2 2

2 2

2

2

 

   





 

















 





 





