Inverse of a Function

Activity - 1 REAL - WORLD ANALOGY (सादृश्य)

Consider two airlines flying between Pokhara and Kathmandu:

Observation:

Yeti Air carries a person from Pokhara to Kathmandu.
Buddha Air carries a person from Kathmandu to Pokhara.
Buddha Air acts opposite to Yeti Air.
Whatever work is done by Yeti Air is undone by Buddha Air.

$(\text{Buddha} \circ \text{Yeti})(\text{Pokhara})$

= $\text{Buddha}\bigl(\text{Yeti}(\text{Pokhara})\bigr)$

= $\text{Buddha}(\text{Kathmandu})$

= $\text{Pokhara}$

$\therefore\; (\text{Buddha} \circ \text{Yeti})(\text{Pokhara}) = \text{Pokhara}$ — Identity function

Activity - 2 (FUNCTION MACHINE)

Consider two function machines with input $x = 3$

Observation:

$f(3) = 4$
$g(4) = 3$
Work/action done by $f$ is undone or reversed by $g$.

$(g \circ f)(3) = g\bigl(f(3)\bigr) = g(4) = 3$

$\therefore\; (g \circ f)(3) = 3$ — Identity function

Conclusion:

Activity 1 Buddha Air acts opposite to Yeti Air, so Buddha Air is the inverse of Yeti Air.

Activity 2 Function $g$ acts opposite to function $f$, so function $g$ is the inverse of function $f$.

Definition:

Let $f: A \to B$ be a function from set $A$ to set $B$. Let $g: B \to A$ be a function defined from set $B$ to $A$.

The function $g$ is said to be inverse of $f$ (or vice-versa) if:

$$f \circ g(x) = g \circ f(x) = x \quad \text{for all } x \in \mathbb{R}$$

It is denoted as: $g^{-1} = f$ or $f^{-1} = g$

Important Note:

📌 For the existence of an inverse, the function must be one-to-one onto (bijective).

Worked Examples

1. Find $f^{-1}$ if $f = \{(a,1),\,(b,2),\,(c,3)\}$

Solution: To find the inverse, swap every ordered pair $(x, y) \to (y, x)$:

$$f^{-1} = \{(1,a),\,(2,b),\,(3,c)\}$$

2. Find $f^{-1}(x)$ if $f(x) = 2x + 3$

Solution:

Step (I) Let $y = f(x)$ Let $f(x) = y$, so $y = 2x + 3$

Step (II) Interchange variables $x$ and $y$

Replace every $x$ with $y$ and every $y$ with $x$:

or, $x = 2y + 3$

Step (III) Make $y$ alone (solve for $y$)

or, $x - 3 = 2y$

or, $\frac{x - 3}{2}$ = $y$

Step (IV)

$\therefore$ $f^{-1}(x) = \frac{x-3}{2}$

Verification (Checking)

We verify by computing $f \circ f^{-1}(x)$:

$f \circ f^{-1}(x)$ $= f\!\left[f^{-1}(x)\right]$ $= f\!\left(\dfrac{x-3}{2}\right)$ $= 2 \cdot \dfrac{x-3}{2} + 3$ $= (x - 3) + 3$ $= x$

$\therefore\; f \circ f^{-1}(x) = x$ — Identity function confirmed.

Why does this work?

Let $y = f(x)$. When we interchange $x$ and $y$:

$x = f(y)$

Taking composite of $f^{-1}$ on both sides:

$f^{-1}(x) = f^{-1}\bigl(f(y)\bigr)$ = ($f^{-1}\circ f$)($y$) = $y$

[Because the composite of a function and its inverse is the identity function.]