The Algebra of Functions

Functions are similar to other algebraic expressions since we can perform all the usual algebraic operations on them. They can be added, subtracted, multiplied and divided, used as exponents etc. There are however restrictions which we must respect, not only due to algebraic properties (such as not dividing by 0) -- there are added restrictions dictated by the domain of the function in question. We can't divide one function by another over all values of x if the divisor function has a zero somewhere. We also can't add, subtract or multiply functions which don't have common elements in their domains.

For example, say we want to add functions f and g where and g(x) = x.

We know that x = 2 is not an element in the domain of f since it makes the denominator zero, therefore, we can't add the two functions at x = 2 because f (2) is undefined. We can add them for every other x-value in the known universe.

 If x is in the domain of functions f and g, we can form new functions using algebraic combinations such as their sum, difference, product and/or quotient. The sum (f + g)(x) = f (x) + g(x) The difference (f – g)(x) = f (x) – g (x) The product fg(x) = f (x) × g(x) The quotient ( f / g )(x) = f (x) / g(x), where g(x) ! 0.

.

. Example 1

If f (x) = 5 – 2x and find (f + g)(2) and (f + g)(– 4)

Solution  but is not a real number, therefore (f + g)(– 4) does not exist in R.

From this example, we see that if one of the functions has a restricted domain, then those restrictions must apply to the domain of the combined function.

The domain of g is ,

therefore ( f + g ) has the same domain restrictions.

Example 2

If f (x) = 8 – 3x and ,find:

 a) (f + g)(4) b) (f – g)(3) c) f × g(7) d) (f /g)(5) e) (gg)(0) f) (g /f)(2)

Solution

a) .

b) .

c) .

d) e) (gg)(0) is undefined; the domain of g is Since 0 is less than 4/3, (gg)(0) doesn't exist.

f) When we need to find a general expression for a combination of functions such as
(f + g)(x), if any of the original functions has domain restrictions, we must apply these restrictions to the domain of the combined function.

So, for part (e) of example 2, had we been asked to find (gg)(x),

our solution would have been Though the combined function 3x – 4 is defined over all real numbers,

the domain restrictions of the original function must apply to the combined function.

Example 3

Given f (x) = x ² + 3 and :

find these combinations of f and g and state the domain:

 a) (f + g)(x) b) (f – g)(x) c) (f · g)(x) d) (f /g)(x) e) ( f · f )(x) f) (g /f )(x)

Solution

a) b) c) d) e) ( f · f )(x) = (x2 + 3)(x2 + 3) = x4 + 6x2 + 9; .

f) .

. In the real world, many functions are composite or combinations of functions. That is, the dependent variable or output of one function becomes the independent variable or input of another function in an ordered sequence.

The salaries which a manufacturer pays employees is often a function of the profits the company makes. The profits in turn are a function of the level of production and the sales, which are of course a function of the promotion, inventory and worker output. So, the output of the sales or production function becomes the input of the salary function and so on.

When we compose or combine a series of functions,

the y-values of the 1st function we apply become the x-values of the 2nd function, etc.

Women's shoe sizes in North America are different from those in Continental Europe which are of course, different from those in Britain. We couldn't keep things simple, now could we?

The function f (x) = 2(x + 12) converts American shoe sizes to European sizes, and,
g(
x) = ½ x 14 converts European shoe sizes to British.

So, say Fancy Shmancy, who wears size 6 shoes in America, wants to buy a pair of shoes in Britain. In order to know what size to ask for, she would have to apply the two conversion functions.

First, she would substitute x = 6 (her American shoe size -- shh!) into f (x) to get 36, which is her shoe size in Europe. Next, she would substitute x = 36 into g(x) to get 4 which is her shoe size in Britain since:

f (6) = 2(6 + 12) = 36 and g(36) = ½(36) 14 = 4

Should Fancy decide to buy a pair of Italian shoes, she would ask for size 36 -- but she would do it very quietly..

The mathematical symbols for such composite functions are f o g(x) and f [g(x)],
pronounced f of g of
x.

The correct order of operations is the inner function first and then the outer function.
So, in our shoe size conversion we found g[ f (6)] since we found f (6) before we found g(36). Had we applied the functions in the wrong order -- poor Fancy would wear size (
11) shoes in Britain and size 2 in Italy -- ouch! I sure wouldn't want to try on a pair of negative sized shoes. I think they'd be hard to find and impossible to lace they'd be so small.

Example 4

If f (x) = 2x + 3 and g(x) = 3(x 2), find:

 a) f [g (1)] b) g [f (3)] c) g [g (2)] d) f [ f (– 2)]

Solution

a) f [g (1)] = f [ 3(1 2)] =

f (3) = 2(3) + 3 = 3

b) g [ f (3)] = g [ 2(3) + 3] =

g (9) = 3 (9 2) = 21

c) g [ g (2)] = g [ 3(2 2)] =

g (0) = 3 (0 2) = 6

d) f [ f ( 2)] = f [2 ( 2) + 3] =

f ( 1) = 2(1) + 3 = 1

. .

1) Given f (x) = x2 + 5 and g(x) = 2x 1; find:

 a) (f + g)(x) b) (g + g)(3) c) (g – f)(x) d) (g + f )(– 1) e) (f g)(2) f) (g /f )(– 2)

2) Given f (x) = 3x 7 and g(x) = x 2 9; find and state the domain of :

 a) f [g(x)] b) g [f (x)] c) f [f (x)] d) g [g (x)] e) (f /g)(x) f) (g /f )(x)

.

. 1) Given f (x) = x2 + 5 and g(x) = 2x 1; find:

a) (f + g)(x) = x 2 + 5 + ( 2x 1) = x 2 2x + 4
b) (g + g)(3) = [ 2(3) 1 ] + [ 2(3) 1 ] = 14
c) (g f )(x) = 2x 1 (x 2 + 5) = x 2 2x 6
d) g [g (x)] = g ( 2x 1) = 2 ( 2x 1) 1 = 4x + 1
e) f) 2) Given f (x) = 3x 7 and g(x) = x 2 9; find and state the domain of :

a) f [g(x)] = f [ x2 9 ] = 3 [ x2 9 ] 7 = 3x2 34 ; domain is R.

b) g [ f (x)] = g (3x 7) = (3x 7) 2 9 = 9x 2 42x + 40 ; domain is R.

c) f [f (x)] = f (3x 7) = 3(3x 7) 7 = 9x 28 ; domain is R.

d) g [g (x)] = g (x2 9) = (x2 9) 2 9 = x 4 18x 2 + 72 ; domain is R.

e) f) .

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