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

   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.

1. (f+g) (x) = f(x) + g(x)        Domain = A∩ B
2. (f-g) (x) = f(x) - g(x)          Domain = A∩B
3. (f*g) (x) = f(x) * g(x)        Domain = A∩B
4. f/g (x) = f(x)/g(x)             Domain = { x | x ∈A∩B  g(x) ≠0}
5. (f+k) (x) = f(x) + k            k is constant, Domain = A
6. f ⁿ(x) = [f(x)]ⁿ                    is a positive integer, Domain = A
7. 
   Domain is a subset of A or A∩{x | f(x) ≥0}
8. | 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) = x^{2-5, g(x) = 1-x}

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

      = x^{2 - 5 + 1 - x}

      = x^{2 - 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

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)
4. 

Solutions:

1. (f+g) (-3) = f(-3) + g(-3)
   = (-3)2 + 1 + 3 * (-3)
   = 9 + 1 - 9
   = 1


2. 


3. (f • g ) (- √2 ) = f (-√2) *g(-√2)
   = [(- √2 )2 + 1] [3*-√2]
   =[2+1] (-3√2)
   =3-(-3√2)
   =-9√2


4. 

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.    f²(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.   f³
     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.   
            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

Example 6