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Algebraic Function
A function is said to be Algebraic if it arises as a result of performing a finite number of fundamental algebraic operations (addition, subtraction, multiplication, division, root extraction, etc.) on polynomial functions.
 

Example
 
http://www.redcomet.org/Ma25b/i64.gif

is an algebraic function.
 
If f and g are real functions with domain sets A, B, respectively, then both f and g are defined on A∩B.

We now define the functions

f+g, f-g, fg, f/g, f+k, kf, f n √f, | f | as follows.               
 
i.
 (f+g) (x) = f(x) + g(x)      Domain = A∩ B
     
ii.
 (f-g) (x) = f(x) - g(x)       Domain = A∩B
     
iii.
 (f*g) (x) = f(x) * g(x)     Domain = A∩B
     
iv.
 f/g (x) = f(x)/g(x)           Domain = { x | x ∈A∩B

                         g(x) ≠0}
     
v.
 (f+k) (x) = f(x) + k         k is constant, Domain = A
   
vi.
 f n(x) = [f(x)]n is a positive integer, Domain = A
   

 viii.

http://www.redcomet.org/Ma25b/i65.gif

 
  Domain is a subset of A or A∩{x | f(x) ≥0}  
     
 ix.
 | f | (x) = |f(x)|            
   Domain = A
     
  The codomain of each of these functions is a real number set.  
     

Example 1
  f: R→R g:R→R defined by f(x) = x2-5, g(x) = 1-x

(f+g) (x) = f(x) + g(x)

            = x2 - 5 + 1 - x

            = x2 - x - 4

Example 2

  f: R→R g:R→R defined by f(x) = 2x - 7, g(x) = 3 + 5x

(f-g) (x) = f(x) - g(x)

           = 2x - 7 - (3+5x)

           = 2x - 7 - 3 - 5x

           = -10 - 3x

Example 3

  f: R→R g: R→R defined by f(x) = 2x2 - 3, g(x) = 1 + x

(f *g) (x) = f(x) * g(x)

              = (2x2 - 3) (1+ x)

              = 2x2 - 3 + (2x2 - 3) x

              = 2x2- 3 + 2x3- 3x

              = 2x3 + 2x2 - 3x - 3

Example 4

  f: R→R g: R→R defined by f(x) = 7x + 9, g(x) = x - 5
where x ≠ 5
   
  http://www.redcomet.org/Ma25b/i66.gif

Example 5

 

Let f: R→R and g: R→R defined by f(x) = x2 + 1 and g(x) = 3x. Find the value of

   
 
  1. (f+g) (-3)

  2. (f-g) (1/4)

  3. (f-g) (-√2)
   
 

   iv.

 http://www.redcomet.org/Ma25b/i68.gif
   
Solutions:
   
i.
 (f+g) (-3) = f(-3) + g(-3)

                = (-3)2 + 1 + 3 * (-3)

                = 9 + 1 - 9

                = 1
 
ii.
 http://www.redcomet.org/Ma25b/i69.gif
 
iii.
 (f • g ) (- √2 ) = f (-√2) *g(-√2)

         = [(- √2 )2 + 1] [3*-√2]
                  =[2+1] (-3√2)
                  =3-(-3√2)
                  =-9√2
 
iv.

http://www.redcomet.org/Ma25b/i71.gif

http://www.redcomet.org/Ma25b/i72.gif

Example 6

  If f(x) = ax4 + bx2 + c is an even function, find c.

For f(x) to be an even function, f(-x) = f(x)

a(-x)4 + b(-x)2 + c= ax4 + bx2 + c

⇒ c can be any real number.

Example 7

  If f(x) = px3+ qx + r is an odd function, find r.

For f(x) to be an odd function, f(-x) = -f(x)

p(-x)3 + q(-x) + r = - (px3 + qx + r)

                ⇒   r = -r

                ⇒ 2r = 0

                ⇒   r = 0

Try these questions
   
1.
 If f(x) = x2 for x < 0

        = x for 0 < x < 1

        = 1/x for x ≥1

then find
   
 
  1. (3f-6) (x)

  2. f 2(3)

  3. f(-6) + f(1/2) - f(2)
   
  Solution :
   
 

i.

(3f - 6) (x) = (3f) (x) - 6

               = 3 f(x) - 6

               = 3x2 - 6 if x < 0

               = 3x - 6 if 0 < x < 1

               = 3/x - 6 if x ≥1
   
 
   
ii.
 f 2(3) = [f(3)]2 = (1/3)2 = 1/9 since 3 > 1
 
 
iii.
 f(-6) + f(1/2) - f(2)

  f(-6) = (-6)2 since -6<0

        = 36

f(1/2) = 1/2 since 0 < 1/2 < 1

   f(2) = 1/2 since 2 > 1

then

f(-6) + f(1/2) - f(2) = 36 + 1/2 - 1/2

                         = 36
 
   
2. f and g are real functions defined by f(x) = 3x + 1 and
g(x) = 4x2. Find
   
 
  1. -5f

  2. f 3

  3. (3f-2g) (x)

  4. (2fg) (x)

  5. g3(x+1)

  6. (√f /g) (x)

  7. (f+g+2) (x)
   
  Solution :
   
 
i.
 (-5f) (x) = -5 (f(x))

           = -5(3x+1)

           = -15x - 5
 
ii.
 f 3(x) = [f(x)]3

        = (3x+1)3
 
iii.
 (3f-2g)(x) = (3f)(x) - (2g)(x)

              = 3f(x) - 2g(x)

              = 3(3x+1) - 2(4x2)

              = 9x + 3 - 8x2

              = 3 + 9x - 8x2
 
iv.
 (2fg)(x) = 2(fg)(x)

           = 2f(x) * g(x)

          = 2(3x + 1) (4x2)

          = 8x2 (3x + 1)

          = 24x3 + 8x2
 
v.
 g3 (x+1) = [g(x+1)]3

             = [4(x+1)2]3

             = 64(x+1)6
 
vi.

http://www.redcomet.org/Ma25b/i73.gif

assuming x≠0

 
vii.
 (f+g+2)(x) = f(x)+g(x)+2

                        = 3x + 1 + 4x2 + 2

                        = 4x2 + 3x + 3
   
 
3.
 Let f: R→Z defined by f(x) = [x]

Find
   
 
  1. f(-4)

  2. f(3)

  3. f(-7/4)

  4. f(5/8)
   
  Solution:
   
 
i.
 f(x) = [x] = n if n ≤ x < n + 1 where x is an integer.

   f(-4) = -4 since -4 ≤ n < -3
 
ii.
 f(3) = [3] = 3 when 3 ≤ 3 n < 4
 
iii.
 f(-7/4) = [-1 3/4] since -7/4 = -1 3/4 , -2 ≤-1 3/4 < -1

                                       = - 2
 
iv.
 f(5/8) = [5/8]

        = 0 since 0 ≤ 5/8 < 1
 

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