# 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) = 2x^{2}- 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/(x^{2}– 4).Find the domain.**Solution:**Let f(x) = 0 then 3x/x

^{2}- 4 = 0 Put (x=2) 3(2)/2^{2}- 4 = 0 6/(4-4) = 0 Put (x= -2) 3(-2)/2^{2}- 4 = 0 6/(4-4) = 0Here, 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.