Functions

Recall

Relation

• Let A and B be two sets. A relation between A and B is a collection of ordered pairs (a, b) such that a A and b B.

Function – Domain – Range

• Let A and B be two sets. A function f from A to B is a relation between A and B such that for each a A there is one and only one associated b B.

• The set A is called the domain of the function, B is called its range.

Examples• Let A = {1, 2, 3, 4}, B = {14, 7, 234},

C = {a, b, c}, and R = real numbers.

Given the following relations, determine if they are functions:

1. r is the relation between A and B that relates the pairs {1 ~ 234, 2 ~ 7, 3 ~ 14, 4 ~ 234, 2 ~ 234}

2. f is the relation between A and C that relates the pairs {(1,c), (2,b), (3,a), (4,b)}

1. g is the relation between A and C consisting of the relates {(1,a), (2,a), (3,a)}

2. h is the relation between R and itself consisting of pairs {(x,sin(x))}

Let A = {1, 2, 3, 4}, B = {14, 7, 234}, C = {a, b, c}, and R = real numbers.

1. r is the relation between A and B that associates the pairs 1 ~ 234, 2 ~ 7, 3 ~ 14, 4 ~ 234, 2 ~ 234

The relation r is not a function, because the element 2 from the set A is associated with two elements from B.

2. f is the relation between A and C that relates the pairs {(1,c), (2,b), (3,a), (4,b)}

The relation f is a function, because every element from A has exactly one relation from the set C.

NO…….WHY?

3. g is the relation between A and C consisting of the associations {(1,a), (2,a), (3,a)}

The relation g is not a function, because the element {4} from the domain A has no element associated with it.

Let A and B be two sets. A function f from A to B is a relation between A and B such that for each a A there is one and only one associated b B.

A = {1, 2, 3, 4}, B = {14, 7, 234},

4. h is the relation between R and itself consisting of pairs {(x,sin(x))}

The relation h is a function with domain R, because every element {x} in R has exactly one element {sin(x)} associated with it.

More ExamplesThis is a function. You can tell by tracing from each x to each y. There is only one y for each x; there is only one arrow coming from each x.

This is a function! There is only one arrow coming from each x; there is only one y for each x. It just so happens that it's always the same y for each x, but it is only that one y. So this is a function; it's just an extremely boring function!

This one is NOT a function: there are two arrows coming from the number 1; the number 1 is associated with two different range elements. So this is a relation, but it is not a function.

Each element of the domain that has a pair in the range. But what about that 16? It is in the domain, but it has no range element that corresponds to it! This won't work! So then this is not a function.

One to One - Onto

A function f from A to B is called one to one (or one - one):

if whenever f(a) = f(b) then a = b.

A function f from A to B is called onto: if for all b in B there is an a in A such

that f(a) = b.

A function f from A to B is called one-to-one (or 1-1) if whenever f (a) = f (b) then a = b. No element of B is the image of more than one

element in A.

One-to-OneNot

One-to-One

A function f from A to B is called one to one (or one - one): if whenever f(a) = f(b) then a = b.

One – to - One

• If a horizontal line intersects the graph of a function in more than one place, then there are two different points a and b for which f(a) = f(b), but a b. Then the function is not one-to-one.

A function f from A to B is called onto if for all b in B there is an a in A such that f (a) = b. All

elements in B are used.

"Onto"(all elements in B are used)

NOT "Onto"(the 8 and 1 in Set B are not

used)

A function f from A to B is called onto:

For all b in B there is an a in A such that f(a) = b.

Onto

If you place a light on the left and on the right hand side of the coordinate system, then the shadow of the graph on the y axis is the image of the domain of the function. If that shadow covers the range of the function, then the function is onto.

Note that every function can be modified to be onto by setting its range to be the image of its domain.

Examples

• Which of the following functions are one-one, onto ? The domain and range for all functions is R.

1) f(x) = 2x + 5 2) g(x) = sin(x) 3) h(x) = 2x3 + 5x2 - 7x + 6

Linear Function

This function is linear. The equation y = 2x + 5 has a unique solution for every x, so that the function is one-one and onto. In fact, all linear functions are one to one and onto.

Periodic Function

Since this function is periodic, it can not be one-to-one. In fact, every periodic function is not one-to-one.

It is onto, if the range is the interval [-1, 1], but not onto if the range is R.

Odd Degree PolynomialThis is an odd-degree polynomial. Hence, the limit as x approaches plus or minus infinity must be plus or minus infinity, respectively. That means that the function is onto. In fact, every odd-degree polynomial is onto while no even degree polynomial is onto.

Since most third degree equations have three zeroes, this function is probably not one-to-one. A look at the graph confirms this.

One – to – One ProofsDefinition - A function f from a set A to a set B is one to one if for

any p and q in A, if f(p) = f(q) then p = q.

Example 1:Prove the function f(x) = 3x + 5 from the real numbers to the real numbers is a one to one function.

Assume f(p) = f(q) then 3p + 5 = 3q + 5 - 5 - 5 3p = 3q thus p = q. Therefore, f(x) is one – to – one.

Example 2:Prove the function f(x) = x2 - 5x + 6 is not one to

one.Since

f(1) = 1 - 5 + 6 = 2 and f(4) = 16 - 20 + 6 = 2 f(1) = f(4)

But, 14 Therefore, f(x) is NOT one – to – one.

• Example 3Prove the function f(x) = x3 +x - 5 from the real numbers to the real numbers is a one to one function.

-How can we show that this is 1-1?

-What do we know about this function?

-Is it always increasing?

-Can we show that?

• Example 3Prove the function f(x) = x3 +x - 5 from the real numbers to the real numbers is a one to one function.

f’(x) = 3x2+1 > 0 for all x. Therefore f(x) is always increasing.Assume p<q , then f(p)<f(q) ….and f(p)f(q) .Therefore f is one-to-one.

Citations

http://web01.shu.edu/projects/reals/index.htmlhttp://regentsprep.org/