Here are some guidelines for determining whether a relation is a function or not.
Function notation is used to name functions for easy reference. Imagine if every function in the world had to start off with y=. Pretty soon, you would become confused about which y= you were talking about. This is the way antecedents work in English. If you just say "it" is red, you really don't have any clue which "it" you're talking about. You need some other way of naming things. Enter function notation.
In this example, the f is a function of x. That is, x is the independent variable, and the value of f depends on what x is. Also, g is a function of both x and y.
The notation f(x) does not mean f times x. It means the "value of f evaluated at x" or "value of f at x" or simply "f of x"
f(3) = 3(3) + 2 = 9 + 2 = 11
f(3) does not mean f times 3. It means the "value of f evaluated when x is 3".
f(t) = 3(t) + 2 = 3t + 2
Whatever is in parentheses on the left side of the function (t in this case) is substituted for the value of the independent variable on the right side.
f(x+h) = 3(x+h) + 2 = 3x + 3h + 2
Every occurrence of the independent variable is replaced by the quantity in parentheses. A common mistake is to take a quantity and apply linear or affine transformations to it.
You also specify which function you want to use when you use function notation. Consider
g(2,1) = 22 + 3(1) = 4 + 3 = 7
Since the order of the independent variables in the original definition was x and then y, the function g is evaluated when x=2 and y=1.
g(1,2) is something completely different. In this case, x=1 and y=2.
g(1,2) = 12 + 3(2) = 1 + 6 = 7
Okay, so it's not completely different, but it is found in a different manner.
Sometimes, functions are a little more complex than the simple functions we've described so far. If different rules are used for different values of the independent variable, then we can use a piecewise definition.
Consider the function shown and the following evaluations.
f(2.1) = 2(2.1) = 4.2
Since 2.1 is in the interval [1,3), we use the second piece of the definition, f(x) = 2x.
f(-2) = 3 - (-2)2 =3 - 4 = -1
Since -2 is in the interval (-∞,-1), we use the first piece of the definition, f(x) = 3-x2
f(3) = 5 - (3) = 5 - 3 = 2
Since 3 is not included in the second piece, but it is included in the third piece on the interval [3,+∞) we use the third piece of the definition, f(x) = 5 - x.
f(0) is undefined
Since 0 doesn't fall into any of the domains, the function is undefined there.
When all of the domains are combined, the domain of the function f, is the set of all real numbers except [-1,1). You could also write it as (-∞,-1)U[1,+∞). The symbol U stands for the union of two sets.
You can put piecewise functions into the graphing calculator. The TI82 and TI83 calculators have a [Test] key obtained by pressing [2nd] [Math]. Under this key, you will find the different test operators (equal to, not equal to, greater than, greater than or equal to, less than, and less than or equal to).
When the calculator evaluates a test expression (x<1), it will return the value of 1 if the statement is true and the value of 0 if the statement is false. This works really well with multiplication, because multiplying by 1 won't change the expression, and multiplying by 0 will make the expression 0.
y = (1 - x2)*(x<-1)
Consider just the first part of the piecewise definition from above. If x<-1, then the calculator will return (1-x^2). If x is not less than negative one, then the calculator will return 0. That's not the proper value to return (it should be undefined), but it will graph on the x-axis and won't show up because the axis is already there. So, it has the semblance on graphing correctly.
The complete piecewise function can be defined for the calculator as:
y = (1-x2)*(x<-1) + 5x*(1≤x and x<3) + (5-3x)*(x≥3)
The "and" keyword can be found under [Test] Logic menu. Similarly, the ≤ and ≥ symbols can also be found under the [Test] menu. If you try to put in 1≤x<3 like we right, the calculator will interpret that as 1≤3 and always return true.
Be sure to use some kind of decimal setting or dot mode when you're viewing a piecewise function. Otherwise, you may get some weird results.
We defined implied domain earlier. It is the set of all real numbers where the expression is defined.
You do not need to state restrictions which result from the implied domain. In other words, if there is an (x-2) in the denominator, you do not need to state that x cannot be 2.
Sometimes, you will need to exclude other values.
If you ever simplify a function, and a value which was in the implied domain is no longer in the implied domain, then it needs to become a stated restriction. Example, divide (x2-4) by (x-2). The implied domain is x cannot equal 2 because that would cause division by zero. However, if you factor the numerator as (x-2)(x+2), then (x-2) in the numerator divides out with the (x-2) in the denominator and you're left with just (x+2). The fact that x cannot be 2 is no longer implied by a simple x+2 in the numerator, so you must now state that x cannot be 2.
When you have a function which is a composition of several pieces, the values in the domain must be able to be used in all parts of the function. Let's say the numerator has a square root of x in it (so that x must be non-negative) and the denominator has an x-2 in it (so that x cannot be 2). When you combine those domains, you get all values of x that are non-negative except 2.
In other words, the domain for the function is the intersection of all the domains of the individual parts.
Now, let's say the numerator has a square root of x in it (so that x must be non-negative) and the denominator has an x+2 in it (so that x cannot be -2). When you combine those domains, you get all values of x that are non-negative except -2. Well, you don't need to state that x cannot be -2 since you've already said that x is non-negative. So, in this case, the domain would be all non-negative values of x.