## 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 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) = 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 #### 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. 5. #### 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 when 3 ≤ 3 n < 4
iii.   f(-7/4) = [-1 3/4] since -7/4 = -1 3/4 , -2 ≤-1 3/4 < -1
= - 2
4.   iv.   f(5/8) = [5/8]          = 0 since 0 ≤ 5/8 < 1