Understanding Special Functions in Mathematics: Constant, Identity, and Real-Valued Functions

Special Cases of Function

Special cases of function

There are some special cases of a function which will be very useful. We discuss some of them below:

• (i) Constant function
• (ii) Identity function
• (iii) Real – valued function

(i) Constant function

A function \(f : A \to B\) is called a constant function if the range of \(f\) contains only one element. That is, \(f(x) = c\), for all \(x \in A\) and for some fixed \(c \in B\).

Illustration 16

From Fig.1.37, \(A = \{a,b,c,d\}\) and \(B = \{1, 2, 3\}\) and \(f = \{(a, 3),(b, 3),(c, 3),(d, 3)\}\). Since, \(f(x) = 3\) for every \(x \in A\), Range of \(f = \{3\}\), \(f\) is a constant function.

(ii) Identity function

Let \(A\) be a non–empty set. Then the function \(f: A \to A\) defined by \(f(x) = x\) for all \(x \in A\) is called an identity function on \(A\) and is denoted by \(I_A\).

Illustration 17

If \(A = \{a,b,c\}\) then \(f=I_A = \{(a,a),(b,b),(c,c)\}\) is an identity function on \(A\).

(iii) Real valued function

A function \(f: A \to B\) is called a real valued function if the range of \(f\) is a subset of the set of all real numbers \(\mathbb{R}\). That is, \(f(A) \subseteq \mathbb{R}\).

Example 1.17

Let \(f\) be a function from \(\mathbb{R}\) to \(\mathbb{R}\) defined by \(f(x) = 3x - 5\). Find the values of \(a\) and \(b\) given that \((a, 4)\) and \((1, b)\) belong to \(f\).

Solution

\(f(x) = 3x – 5\) can be written as \(f = \{(x, 3x – 5) | x \in \mathbb{R}\}\)

\((a, 4)\) means the image of \(a\) is 4. That is, \(f(a) = 4\)

\(3a – 5 = 4 \implies a = 3\)

\((1, b)\) means the image of 1 is \(b\).

That is, \(f(1) = b \implies 3(1) – 5 = b \implies b = -2\)

Example 1.18

The distance \(S\) (in kms) travelled by a particle in time ‘\(t\)’ hours is given by \(S(t) = \frac{t^2 + t}{2}\). Find the distance travelled by the particle after

(i) three and half hours.

(ii) eight hours and fifteen minutes.

Solution

The distance travelled by the particle in time \(t\) hours is given by \(S(t) = \frac{t^2 + t}{2}\).

(i) \(t = 3.5\) hours. Therefore, the distance travelled in 3.5 hours is 7.875 kms.

(ii) \(t = 8.25\) hours. Therefore, the distance travelled in 8.25 hours is 38.16 kms, approximately.

Example 1.19

If the function \(f : \mathbb{R} \to \mathbb{R}\) is defined by:

then find the values of:

(i) \(f(4)\)

(ii) \(f(-2)\)

(iii) \(f(4) + 2f(1)\)

(iv) \(\frac{f(1) + 3f(4)}{f(-3)}\)

Solution

The function \(f\) is defined by three values in intervals I, II, III as shown by the side:

For a given value of \(x = a\), find out the interval at which the point \(a\) is located, there after find \(f(a)\) using the particular value defined in that interval.

(i) First, we see that, \(x = 4\) lie in the third interval.
Therefore, \(f(x) = 3x - 2\); \(f(4) = 3(4) – 2 = 10\)

(ii) \(x = -2\) lies in the second interval.
Therefore, \(f(x) = x^2 – 2\); \(f(-2) = (-2)^2 – 2 = 2\)

(iii) From (i), \(f(4) = 10\).
To find \(f(1)\), first we see that \(x = 1\) lies in the second interval.
Therefore, \(f(x) = x^2 – 2 \implies f(1) = 1^2 – 2 = -1\)
So, \(f(4) + 2f(1) = 10 + 2(-1) = 8\)

(iv) We know that \(f(1) = -1\) and \(f(4) = 10\).
For finding \(f(-3)\), we see that \(x = -3\), lies in the first interval.
Therefore, \(f(x) = 2x + 7\); thus, \(f(-3) = 2(-3) + 7 = 1\)

Hence,