Problem 1: Find the domain of the function given by :

f x = x [ x − 2 ] [ x + 1 ] f x = x [ x − 2 ] [ x + 1 ]

Solution :

Statement of the problem : The function has rational form. Denominator consists of product of two greatest integer functions.

We can consider, the function as product of three individual functions :

f x = x X 1 [ x − 2 ] X 1 [ x + 1 ] f x = x X 1 [ x − 2 ] X 1 [ x + 1 ]

The domain of "x" is “R”. We, now, analyze individual greatest integer functions such that it does not become zero. If we recall the graph of greatest integer function, then we can realize that the value of greatest integer [x] is equal to zero for the interval given by 0≤ x < 1. Following this clue, we find the intervals in which greatest integer functions are zero.

Figure 1: The greatest integer function evaluates to zero for 0≤ x < 1.

Greatest integer function

For [ x − 2 ] = 0 [ x − 2 ] = 0 ,

[ x − 2 ] = 0, if 0 ≤ x − 2 < 1 [ x − 2 ] = 0, if 0 ≤ x − 2 < 1

⇒ [ x − 2 ] = 0, if 2 ≤ x < 3 ⇒ [ x − 2 ] = 0, if 2 ≤ x < 3

⇒ [ x − 2 ] = 0, if x ∈ [ 2,3 ) ⇒ [ x − 2 ] = 0, if x ∈ [ 2,3 )

It means that given function is undefined for this interval of “x”. The domain of the function for this condition is :

D 1 = R − [ 2,3 ) D 1 = R − [ 2,3 )

Similarly, for [ x + 1 ] = 0 [ x + 1 ] = 0

[ x + 1 ] = 0, if 0 ≤ x + 1 < 1 [ x + 1 ] = 0, if 0 ≤ x + 1 < 1

⇒ [ x + 1 ] = 0, if - 1 ≤ x < 0 ⇒ [ x + 1 ] = 0, if - 1 ≤ x < 0

⇒ [ x + 1 ] = 0, if x ∈ [ - 1,0 ) ⇒ [ x + 1 ] = 0, if x ∈ [ - 1,0 )

The domains for this condition is :

D 2 = R − [ - 1,0 ) D 2 = R − [ - 1,0 )

Hence, domain of the given function is intersection of two domains as shown in the figure. Note that we have not considered the domain of numerator, “x”, as its domain is “R” and its intersection with any interval is interval itself.

Figure 2: The domain of the function is intersection of domains of individual functions.

Domain of the function

Domain = D 1 ∩ D 2 Domain = D 1 ∩ D 2

Domain = - ∞ , - 1 ∪ [ 0,2 ) ∪ [ 3, ∞ ) Domain = - ∞ , - 1 ∪ [ 0,2 ) ∪ [ 3, ∞ )

Problem 2: Find the domain of the function given by :

f x = log e cos x − 5 + 9 − x 2 f x = log e cos x − 5 + 9 − x 2

Solution :

Statement of the problem : The given function is sum of logarithmic and algebraic function.

Here, we observe that argument (input) of logarithmic function is itself a trigonometric function. We know that cosine function is real for all real values of "x". The important point to realize here is that we have to evaluate logarithmic function for the values of trigonometric function "cos(x-5)" – not for independent variable “x”. Now, the argument of logarithmic function is a positive number. It means that :

cos x − 5 > 0 cos x − 5 > 0

The basic interval of cosine function is [ - π / 2, π / 2 ] [ - π / 2, π / 2 ] . The solution of the cosine inequality is the domain of the logarithmic function :

D 1 = 2 n π − π 2 < x − 5 < 2 n π + π 2, n ∈ Z D 1 = 2 n π − π 2 < x − 5 < 2 n π + π 2, n ∈ Z

⇒ D 1 = 2 n - 1 2 π + 5 < x < = 2 n + 1 2 π + 5, n ∈ Z ⇒ D 1 = 2 n - 1 2 π + 5 < x < = 2 n + 1 2 π + 5, n ∈ Z

For the algebraic function, the expression within the square root is non-negative number :

⇒ 9 − x 2 ≥ 0 ⇒ x 2 − 9 ≤ 0 ⇒ x + 3 x - 3 ≤ 0 ⇒ 9 − x 2 ≥ 0 ⇒ x 2 − 9 ≤ 0 ⇒ x + 3 x - 3 ≤ 0

Clearly, roots of the quadratic equation, when equated to zero, is -3,3. Here, coefficient of quadratic equal equation is positive. Therefore, middle section is negative. Hence, its domain is :

D 2 = - 3 ≤ x ≤ 3 D 2 = - 3 ≤ x ≤ 3

The domain of given function, f(x), is intersection of two functions i.e.

⇒ D = D 1 ∩ D 2 ⇒ D = D 1 ∩ D 2

From the figure, the common interval is between - 5 π / 2 and - 3 π / 2 - 5 π / 2 and - 3 π / 2 as obtained for n = -1.

Figure 3: The domain of the function is intersection of domains of individual functions.

Domain of the function

D = - 5 π 2 , - 3 π 2 D = - 5 π 2 , - 3 π 2