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

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

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. |

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algebra of functions | composition of functions | practice | solutions |

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**Example 1**

If *f* (*x*) = 5 – 2** x** and find (

**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.*

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**Example 2**

If *f* (*x*) = 8 – 3*x* 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 3*x* – 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

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*) = (*x*^{2} + 3)(*x*^{2} + 3) = *x*^{4} + 6*x*^{2} + 9; .

f)

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algebra of functions | composition of functions | practice | solutions |

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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

*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(**

pronounced

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

**Example 4**

If *f *(*x***) = 2***x*** + 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

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algebra of functions | composition of functions | practice | solutions |

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1) Given *f ***( x) = **

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***) = 3***x*** **–** 7** and *g***(***x***) = ***x*** ^{2} **–

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

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algebra of functions | composition of functions | practice | solutions |

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1) Given *f ***(***x***) = ***x*** ^{2} + 5** and

- a) (

- b) (

- c) (

- d)

- e)

- f)

2) Given *f ***( x) = 3**

a) *f* [*g*(*x*)] = *f* [ *x*^{2} – 9 ] = 3 [ *x*^{2} – 9 ] – **7** = **3***x*** ^{2} **–

b) *g* [ *f *(*x*)] = *g* (3*x* – 7) = (3*x* – 7)^{ 2} – 9 = **9***x*** ^{ 2} **–

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

d) *g* [*g *(*x*)] = *g* (*x*^{2} – 9) = (*x*^{2} – 9)^{ 2} – 9 = *x*** ^{ 4} **–

e)

f)

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*(all content of the MathRoom Lessons **© Tammy the Tutor; 2004 - ).*