On the functions introduction page, explained was how the domain of a function, is the values that go in to the function.
Sometimes restrictions or conditions have to be put on the numbers that can be in the domain of functions in certain situations.
Generally this means that some values will end up having to be excluded from the domain of a function.
Though before looking at some basic domain of functions examples though, it helps to get familiar with some Interval Notation first.
Domain of Functions Examples
(1.1)
f(x) = \space 2x + 4
This function can work for any real value of x.
For this domain we can say x \isin \reals, or ( –∞ , ∞ ).
(1.2)
f(x) = \space 10x
This function will also work for any real x values.
Again the domain is x \isin \reals, or ( –∞ , ∞ ).
(1.3)
f(x) = {\frac{4}{x \space {\text{--}} \space 5}}
Now here we can’t have values of x = 5 in the domain of f(x).
As this would create a division by 0. Which we can’t do.
For the domain we can write x \isin \reals, \space x \space {\cancel{=}} \space 5. To mean that x is any real number, but not the number 5.
In form of the alternative notation, we have a combination of cases.
( –∞ , 5 ) and ( 5 , ∞ )
The brackets can be put together to represent the domain of a function.
In set notation, we have seen the union symbol U.
The symbol is used t mean one group OR another group.
So to represent group G or H, this would look like G U H.
This U symbol can be used for relevant interval notation also.
So ( –∞ , 5 ) and ( 5 , ∞ ) as a domain can be combined with each other as:
( –∞ , 5 ) U ( 5 , ∞ )
Which says that the domain can be a number smaller than 5, OR greater than 5, but NOT 5.
(1.4)
g(x) = {\frac{3}{x^2 \space {\text{--}} \space 5x + 4}}
In this example like with (1.3), we need to avoid having a divisor that is equal to 0
x^2 - 5x + 4 ≠ 0
Factor:
(x - 1)(x - 4) ≠ 0
x ≠ 4 , x ≠ 1
Domain: x \isin \reals , x ≠ 1 x ≠ 4
or ( –∞ , 1 ) U ( 1 , 4 ) U ( 4 , ∞ )
(1.5)
h(x) \space = \space \sqrt{x - 2}
When we’re working with real numbers. We don’t want a negative number inside a square root.
So x - 2 will have to be larger or equal to 0.
x - 2 > 0 is true if x > 2
Domain: x > 2 or [ 2 , ∞ )
(1.6)
f(x) = {\frac{3}{\sqrt{x \space {\text{--}} \space 2}}}
With this situation, like in example (1.5) the number inside the square root can’t be negative.
But as the root is in the denominator of a fraction, it also can’t be equal to 0.
Thus here, x - 2 has to be greater than 0.
x - 2 > 0 is true if x > 2
Domain: x > 2 or ( 2 , ∞ )