Problem 1 : A function is given by :

f x = 1 x + 1 f x = 1 x + 1

Determine its domain set.

Solution : The function, in the form of rational expression, needs to be checked for its denominator. The denominator should not evaluate to zero as “a/0” form is undefined. For given function in the question, the denominator evaluates to zero when,

x + 1 = 0 x + 1 = 0

⇒ x = - 1 ⇒ x = - 1

Hence, domain of the given function is R − { - 1 } R − { - 1 } . The representation of the domain on real number line is shown with a dark line on either side of the excluded point “-1”.

Figure 1: The number "-1" is excluded from the domain.

Domain of a function

Problem 2 : A function is given by :

f x = x 2 − 5 x + 6 f x = x 2 − 5 x + 6

Determine its domain set.

Solution : We observe that given function is a square root of a quadratic polynomial. The expression within square root should be a non-negative number as square root of a negative number is not a real number. It means that expression under even root should be non-negative.

We factorize the quadratic expression in order to find corresponding interval for which expression under the root is non-negative.

⇒ x 2 - 5 x + 6 ≥ 0 ⇒ x 2 - 5 x + 6 ≥ 0

⇒ x − 2 x − 3 ≥ 0 ⇒ x − 2 x − 3 ≥ 0

We interpret this result in reference to quadratic equation. When x ≤ 2, both factors of quadratic equation are non-positive and their product is non-negative. We can check with one such value like “1” or “-1”. On the other hand, when x ≥ 3, both factors are non-negative and their product is also non-negative. It turns out that these two conditions correspond to two intervals, which are disconnected to each other.

Figure 2: The domain consists of two disjointed intervals.

Domain of a function

The representation of the domain interval on real number line is shown with thick line and small filled circles. From the representation on the figure also, it is clear that it is a case of two disjoint intervals. We, therefore, represent the valid domain of the function with the help of the concept of union of two sets (intervals) in the following manner :

- ∞ < x ≤ 2 or 3 ≤ x < ∞ - ∞ < x ≤ 2 or 3 ≤ x < ∞

⇒ - ∞ < x ≤ 2 ∪ 3 ≤ x < ∞ ⇒ - ∞ < x ≤ 2 ∪ 3 ≤ x < ∞

i.e.

⇒ ( - ∞ , 2 ] ∪ [ 3, ∞ ) ⇒ ( - ∞ , 2 ] ∪ [ 3, ∞ )

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

Problem 3 : A function is given by :

y = 9 − x 2 y = 9 − x 2

Determine its range.

Solution : For real value of “y”, the expression 9 − x 2 9 − x 2 is non-negative number. It means that :

⇒ 9 − x 2 ≥ 0 ⇒ 9 − x 2 ≥ 0

⇒ x 2 − 9 ≤ 0 ⇒ x 2 − 9 ≤ 0

⇒ x − 3 x + 3 ≤ 0 ⇒ x − 3 x + 3 ≤ 0

We interpret this result in reference to the given quadratic equation. When x ≥ - 3, but x ≤ 3 x ≥ - 3, but x ≤ 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.

Figure 3: The domain lies between two end points, inclusive of them.

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 ≤ x ≤ 3 - 3 ≤ x ≤ 3

or

[ - 3,3 ] [ - 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 = 9 − x 2 y = 9 − x 2

⇒ x = ± 9 − y 2 ⇒ x = ± 9 − y 2

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

[ - 3,3 ] [ - 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 = 9 − x 2 y = 9 − x 2

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 :

[ 0,3 ] [ 0,3 ]

Problem 4 : A function is given by :

y = x 2 1 + x 2 y = 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 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 = x 2 1 + x 2 y = x 2 1 + x 2

⇒ y x 2 + y = x 2 ⇒ y x 2 + y = x 2

⇒ x 2 = y 1 − y ⇒ x 2 = y 1 − y

⇒ x = ± y 1 − y ⇒ x = ± y 1 − y

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) :

For the first requirement, the expression in the square root should be greater than or equal to zero i.e non-negative number.

y 1 − y ≥ 0 y 1 − y ≥ 0

⇒ y ≥ 0 ⇒ y ≥ 0

Further, denominator of the expression “1-y” is non-negative. Also, “1-y” is not equal to zero. Combining two requirements, the expression is a positive number :

1 − y > 0 1 − y > 0

⇒ y < 1 ⇒ y < 1

Combining two intervals i.e. intersection of two intervals, we have the range of the function as :

Figure 4: Range of the function is equal to intersection of two intervals.

Range of a function

⇒ 0 ≤ y < 1 ⇒ 0 ≤ y < 1