# GRE Algebra | Functions

An algebraic expression consist one variable is used to define a function of that variable. For example, the expression 2y – 7 can be used to define a function f by,

`f(y) = 2y - 7 `

where f(y) is called the value of f at y and function value can be obtained by putting the value of y in above expression.If y=1, then f(1) is

```f(1) = 2(1) - 7
f(1) = -5 ```

Hence, the value of f(1) is -5.

For each input value of y, the function f(y) gives exactly one output.So, it can be say that function f act as a machine which takes an input of the variable y and produces the corresponding output.

It might be possible sometimes when more than one value of y can give the same output of f(y). For example:

```f(x) = 2x2 - 4x + 7

Put (x = 0)
f(0) = 2(0) - 4(0) + 7
= 7

Put (x = 2)
f(2) = 2(2)2 - 4(2) + 7
= 8 - 8 + 7
= 7 ```

Hence, for both values of x = 0, 2 f(x) gives the same output.

Domain of a function is the set of all the permissible inputs. For example:

```g(y) = √(x-5) is function.
√(x-5) ≥ 0
x ≥ 5 ```

Here, domain of the function is [5, ∞)

• Example-1: Let f be the function defined by f(x)= 3x/(x2 – 4).Find the domain.
Solution:

```Let f(x) = 0 then
3x/x2 - 4 = 0

Put (x=2)
3(2)/22 - 4 = 0
6/(4-4) = 0

Put (x= -2)
3(-2)/22 - 4  = 0
6/(4-4) = 0  ```

Here, f is not defined at (x = 2 and x = -2) because 6/0 is not defined.
Hence, the domain of f consists of all real numbers except 2 and -2.

• Example-2: Let f be the function defined by g(x) = √(x – 3) + 11. Find the domain.
Solution:

`√(x - 3) + 11 = 0 `

In this case if x < 3, then g(x) is not a real number.
Hence, domain of g consists of all the real numbers x such that x ≥ 3.

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