# Analytical Representation of the Function

To define function, we need to specify way, that allows us for every value of argument find corresponding value of the function. The most common way to set function is to represent it with the help of formula y=f(x), where f(x) is some expression with variable x. In this case it is said that function is represented by formula or function is represented analytically.

For example, let y=f(x)=x^2+5x-1, where x>=0. Domain of this function is interval [0,+oo]. To find value of the function at any point x>=0, it is sufficiently to find numerical value of expression x^2+5x-1 at chosen point. For example, y(0)=f(0)=0^2+5*0-1=-1, y(2.1)=f(2.1)=(2.1)^2+5*2.1-1=13.91.

For function, that is represented analytically, we sometimes don't specify domain explicitly. In this case, this means that domain of the function is domain of the expression f(x), i.e. set of those values x, for which expression f(x) makes sense.

Example 1. Find domain of the function y=1/(x+2).

Expression 1/(x+2) is defined for all x, except values, where denominator equals zero, i.e. x+2=0 or x=-2. Therefore, domain of the function is all x, except x=-2.

Example 2. Find domain of the function y=sqrt(x-1).

Since square root of negative number is not defined, then expression sqrt(x-1) will make sense when x-1>=0 or x>=1. Therefore, domain of the function is interval [1,+oo).

Sometimes function can be represented piecewise. This means that on different intervals different formulas are used. For example, f(x)={(2x+3 if -1<=x<0),(x+2 if 0<=x<=1):}.

This function is defined on segment [-1,1]. To find value of the function we just need to know, what formula to use for given value of argument. If we need to calculate f(-0.5), then we use first formula, because -1<=-0.5<0, so f(-0.5)=2*(-0.5)+3=2. If we need to calculate f(0.5), then we use second formula, because 0<=0.5<=1, so f(0.5)=0.5+2=2.5. For f(0) we use second formula because 0 belongs to the second interval: f(0)=0+2=2.