2.08  Algebra of Functions

       Piecewise Functions

College Algebra: One Step at a Time,  Pages 296-302: 7a), b), c), 8a, b), c), Extra, 12

                                       Pages 303-305: 1, 4, 8

Dr. Robert J. Rapalje

Seminole State College of Florida

Sanford, FL  32773

Algebra of Functions -- Composite Functions

NOTE:  In this section, there are TWO very important concepts:

                  1.  Substitution

                  2.  Simplification

The first step is usually not hard, and the second step sometimes is easy.  But then again, the second step can be VERY difficult.  Don’t be intimidated!!

Page 298:   7.   and

Solution:  7a)       Find      

Step 1:  Substitution:         

Step 2:  Simplification:

Page 298:   7.   and

Solution:    7b)   Find

Step 1:  Substitution:         

Step 2:  Simplification:

Before going any further, the LCD is , so multiply numerator and denominator of the second fraction by , and the third “fraction” by .  Hang on!  This one might get ugly!

Page 298:   7.   and

Solution:  7c)   Find  .

Step 1:  Substitution:         

                                            , where

Step 2:  Simplification:

Of course, this is a complex fraction, which can be tricky.  Method I or Method II?  Let’s try Method II.

The LCD for the entire fraction is , so multiply numerator and denominator of the complex fraction by .

Now THAT’S UGLY!!  But it cleans up nicely!  ALL these ugly fractions divide out, leaving this:

Page 298:   8.   and

Solution:     8a)     Find     

Step 1:  Substitution:         

Step 2:  Simplification:

Page 298:   8.   and

Solution:  8b)   Find .

Step 1:  Substitution:          

Step 2:  Simplification:

Before going any further, the LCD is , so multiply numerator and denominator of the last fraction by .  Hang on!  This one might get ugly!

Page 298:   8.   and

Solution:  8c)  Find  .

Step 1:  Substitution:          

                                            , where

Step 2:  Simplification:

Of course, this is a complex fraction, which can be tricky.  Method I or Method II?  Let’s try Method II.

The LCD for the entire fraction is , so multiply numerator and denominator of the complex fraction by .

Now THAT’S UGLY!!  But it cleans up nicely!  ALL these ugly fractions divide out, leaving this:

-----------------------------------------------------------------

Extra Application  (from Soledad at Brevard Community College).
The function to convert Fahrenheit to Celsius is given by:
               C(f) = 5/9 (f - 32)

The function to convert Celsius to Kelvin is given by:
               K(C) = C + 273.15

a)      Find a composite function that represents temperature on the Kelvin scale in terms of degrees Fahrenheit.  That is, find  K o C(f)  

               and

b)     Convert 0 degrees Fahrenheit to Kelvin using your composite function. 

Solution:  If , then

c)    Convert 100 degrees F to Kelvin using your composite function.

Solution:   If , then

-----------------------------------------------------------------

Page 301:   12.           and   

Solution:

First find:               and   

12a)                                    b)  

   c)                                     d)  

   e)                                    f)  

                         where              where

   h)                                     g)                                    

              where                     where                

Piecewise Functions

Page 303: #1.     

Solution: 

a)        This means that x=2, which means that x is in category 1, since x ≥ 0 .

b)     This means that x=−2, which means that x is in category 2, since x ≤ 0 .

c)     This means that x=−8, which means that x is in category 2, since x ≤ 0 .

d)        This means that x=8, which means that x is in category 1, since x ≥ 0 .

e)        This means that x=0, which means that x is in category 1, since x ≥ 0 .

f)   This means that x=−25, which means that x is in category 2, since x ≤ 0 .

Page 304: #4.  

Solution: 

a)        This means that x=2, which means that x is in category 3, since x > 0 .

b)     This means that x=−2, which means that x is in category 1, since x ≤ -2.

c)     This means that x=−8, which means that x is in category 1, since x ≤ -2.

d)        This means that x=8, which means that x is in category 3, since x > 0 .

e)    This means that x=0, which means that x is in category 2, since.

f)   This means that x=−13, which means that x is in category 1, since x ≤ -2 .

Page 305: #8.

Solution: 

a)      This means that x=2, which means that x is in category 3, since  .

b)  This means that x=−3, which means that x is in category 2, since.

c)     This means that x=−5, which means that x is in category 1, since x < -3.

d)        This means that x=8, which means that x is in category 3, since .

e)    This means that x=0, which means that x is in category 2, since.

f)   This means that x=−8, which means that x is in category 1, since x < -3 .

Return to main page        Math in Living C O L O R !!

     Return to Basic Algebra page     Return to Intermediate Algebra page  

 Return to College Algebra page