Domain and range  (Page 3/4)

While determining range of a function, we need to be careful with regard to two important aspects :

1: The values obtained for range are consistent with the function. This means that we should check the range against the requirement of a given function. For example, if range of a square root function is evaluated as say [-3,3], then we need to discard negative interval. A square root can not be negative. Hence, the correct range would be [0,3].

2: We need to exclude values of function (y) corresponding to invalid values of “x”. This is particularly the case if domain is a continuous interval except few points barred by the definition of the function.

The best way to understand this algorithm is to work with few examples.

Examples

Problem 3 : A function is given by :

$$y=\sqrt{\left(9-x^2\right)}$$

Determine its range.

Solution : For real value of “y”, the expression $\left(9-x^2\right)$ is non-negative number. It means that :

$$\Rightarrow 9-x^2\ge 0$$

$$\Rightarrow x^2-9\le 0$$

$$\Rightarrow \left(x-3\right)\left(x+3\right)\le 0$$

We interpret this result in reference to the given quadratic equation. When $x\ge -\mathrm{3,}\text{but}x\le 3$ , the signs of two factors are opposite and hence their product is less than or equal to zero. Outside this interval, the expression evaluates to positive number.

Domain of a function

The representation of the domain interval on real number line is shown with thick line and two small filled circles. We see that real values of “x” lies between “-3” and “3”, including end points. We represent the valid domain as :

$$-3\le x\le 3$$

or

$$[-\mathrm{3,3}]$$

This interpretation is typical of product of two linear factors, which is less than or equal to zero. This interpretation can also be used as an axiom in general for deciding interval, involving two factors.

In order to find range, we solve the function for “x”,

$$y=\sqrt{\left(9-x^2\right)}$$

$$\Rightarrow x=\pm \sqrt{\left(9-y^2\right)}$$

Following the same analysis as for domain, we reach the conclusion that the value of “y” for real “x” is given by the interval :

$$[-\mathrm{3,3}]$$

Now, the examination of given function reveals that “y” can be only positive number (note that no negative sign precede square root expression) in the expression for "Y" :

$$y=\sqrt{\left(9-x^2\right)}$$

Hence, “y” can not be negative. Note that we determined interval of "y", which includes negative numbers also. Thus, we conclude that range of the given function is half of the interval obtained earlier, which includes zero also :

$$\left[\mathrm{0,3}\right]$$

Problem 4 : A function is given by :

$$y=\frac{x^2}{1+x^2}$$

Determine its range.

Solution : We observe that the both numerator and denominator of the given function are non-negative. It is because “ $x^2$ “ always evaluates to non-negative number. It means that given function is real for all values of “x”. Thus, domain of function is “R”.

In order to find range, we solve the function for “x”,

$$y=\frac{x^2}{1+x^2}$$

$$\Rightarrow y{x}^2+y=x^2$$

$$\Rightarrow x^2=\frac{y}{1-y}$$

$$\Rightarrow x=\pm \sqrt{\left(\frac{y}{1-y}\right)}$$

For “x” to be real, the expression within square root should be non - negative. This case, however, is different in that it is a ratio of two linear expressions. It is possible that denominator is positive, but numerator is negative or vice - versa. As such, the rational expression as a whole will be negative. In the nutshell, we need to evaluate “x” for following requirements (Note : we are presenting basic reasoning here. Subsequently, we will learn more sophisticated means to determine valid intervals of variables) :