4. Special Types of Functions

4.1 Function of 2 Sets

A function defined from Set A to a Set B is one wherein every element in A is related to exactly one element in B. Note that more than one element from set A could be related to the same element in Set B.

Example 2

If A ={a, b, c} and B = {p, q, r}, which of the following can be said to be a function of Set A to Set B?

$\mathrm{S}_1$ = {(a, p), (a, q), (b, q), (c, r)}
$\mathrm{S}_2$ = {(a, p), (b, p), (c, r)}
$\mathrm{S}_3$ = {(a, p), (b, p), (c, p)}
$\mathrm{S}_4$ = {(a, p), (b, q), (c, s)}

(1) $\mathrm{S}_1$ only $\space \space \space$ (2) $\mathrm{S}_2$ only $\space \space \space$ (3) $\mathrm{S}_2$ and $\mathrm{S}_1$ only $\space \space \space$ (4) $\mathrm{S}_1$ and $\mathrm{S}_4$ only

Solution

S$\bm{_1}$ is not a function as $a$ is related to 2 values, i.e., $p$ and $q$.

S$\bm{_4}$ is not a function as $s$ is not in the Set B.

S$\bm{_2}$ and S$\bm{_3}$ are functions as each element in Set A is related to exactly one element in Set B.

Answer:(3) S$_2$ and S$_3$ only

4.2 Even and Odd Functions

An Even Function is one where $f(x) = f(-x)$ . When graphically represented, even functions are symmetrical with respect to the y-axis. Examples: $f(x) = \| x \|$ and $g(x) = x^2 + 4$ Note:All algebraic functions where all the powers of $x$ are even will necessarily satisfy $f(x) = f(- x)$ and , therfore, are even functions.
An Odd Function is one where $f(x) = -f(-x)$. When graphically represented, odd functions are symmetrical with respect to the origin. Examples: $f(x) = x^3 - 3x$ and $g(x) = x^5 + x^7$ Note:All algebraic functions where all the powers of $x$ are even will necessarily satisfy $f(x) = - f(- x)$ and, therefore, are odd functions.

4.3 Test for Function (Vertical line)

Any vertical line will pass through the $x$-axis at a certain $x$-value. The points where the vertical line intersects the graph of the expression (or relation) indicate the number of $\bm{f(x)}$ values (outputs) for that $\bm{(x)}$-value (input). A function will have only one $f(x)$ value (output) for any given $x$-value (input). If any vertical line drawn intersects the graph of a relation at more than $\bm{1}$ point, then such a relation is not a function.

Example 3

Which of the following relations are functions?

(i)	(ii)	(iii)
(iv)	(v)	(vi)

Solution

Vertical lines drawn in (i), (ii) and (vi) as shown below intersect the graphs at 2 or more points, i.e., there is more than one output at those $x$-values. Therefore, these are not functions.

(i)

(ii)

(iii)

(iii), (iv) and (v) are functions as any vertical line drawn passes intersects the graph at only $1$ point.

Answer: (iii), (iv) and (v)

4.4 Finding the inputs (Horizontal line)

To find the number of x-values that satisfy $f(x) = k$, where k is a constant, draw a horizontal line passing through $f(x) = k$. The number of points at which the horizontal line intersects with the graph of the function provide the number of solutions. The vertical lines drawn from these intersection points to the x-axis provide the x-values that satisfy $f(x) = k$.

Example 4

How many x-values satisfy f(f(x)) $= 4$?

Solution

In Figure (i), we note that the horizontal line drawn at $(x) = 4$ intersects the graph of the function at $3$ points, where the x-values are approximately $ -7, 1$ and $4.5$ (identified by drawing vertical lines.
∴ $f(-7) = f(1) = f(4.5) = 4$

As $f(f(x)) = 4$, we need to find the number of x-values that satisfy $f(x) = -7, f(x) = 1$ and $f(x) = 4.5$.
In Figure (ii), we once again draw horizontal lines and count the intersection points with the function. $f(x) = 1$ and $f(x) = 4.5$ intersect at $3$ points each, while $f(x) = -7$ intersects at none of the points.
∴ Total intersection points $= 3 + 3 = 6$

Figure (i)

Figure (ii)

Answer: $6$

4.5 Inverse of Functions

To find the inverse of a function, please adhere to the following
1) Replace $f(x)$ with $y$.
2) Express $x$ in terms of $y$.
3) In thee equation, replace $x$ with $f^{-1}(x)$ and $y$ with $x$.

Example 5

If $f(x) = \dfrac{3x + 5}{7}$, then $f^{-1}(5) \space = \space ?$

Solution

Let $y = \dfrac{3x + 5}{7} \implies 7y = 3x + 5$

⇒ $ x = \dfrac{7y - 5}{3}$

⇒ $ f^{-1}(x) = \dfrac{7x - 5}{3}$

$f^{-1}(5) = \dfrac{(7 \times 5) - 5}{3} = 10$

Answer: $10$