Properties of Inverse Function

Property of Composition with Inverse

This property is the core of the inverse function definition: the inverse function "undoes" the effect of the original function, and vice versa. If we compose a function with its inverse (in any order), we get the identity function $I(x) = x$.

1. Composition of $f$ with $f^{-1}$:

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

   This holds for all $x$ in the domain of $f^{-1}$ (which is the range of $f$).

2. Composition of $f^{-1}$ with $f$:

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

   This holds for all $x$ in the domain of $f$.

Example:

We know that if $f(x) = 2x + 3$, its inverse is $f^{-1}(x) = \frac{x - 3}{2}$. Let's verify the composition property:

• $f(f^{-1}(x)) = f\left(\frac{x - 3}{2}\right) = 2\left(\frac{x - 3}{2}\right) + 3 = (x - 3) + 3 = x$
• $f^{-1}(f(x)) = f^{-1}(2x + 3) = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x$

Both compositions result in $x$, as expected.

Property of the Inverse of an Inverse

If we find the inverse of an inverse function, we get back the original function.

This makes sense because the process of finding an inverse is a "reversal". If we reverse something twice, we return to the original state.

Property of the Inverse of a Composition

If we have a composition of two functions, both of which have inverses, the inverse of the composition is the composition of their inverses, but in reverse order.

Let $f$ and $g$ be two functions with inverses $f^{-1}$ and $g^{-1}$. Then the inverse of the composition $f \circ g$ is:

Note the reversed order: $g^{-1}$ is applied first, then $f^{-1}$.

Analogy: Imagine putting on socks ($g$) and then shoes ($f$). To undo this (the inverse), you must take off the shoes ($f^{-1}$) first, then take off the socks ($g^{-1}$). The order is reversed.

Example:

Let $f(x) = x + 1$ (its inverse is $f^{-1}(x) = x - 1$) and $g(x) = 3x$ (its inverse is $g^{-1}(x) = \frac{x}{3}$).

1. Find $(f \circ g)(x)$:

   $$(f \circ g)(x) = f(g(x)) = f(3x) = 3x + 1$$

2. Find the inverse of $(f \circ g)(x)$:

   Let $y = 3x + 1$. Swap $x$ and $y$: $x = 3y + 1$.

   Solve for $y$: $x - 1 = 3y \implies y = \frac{x - 1}{3}$.

   So, $(f \circ g)^{-1}(x) = \frac{x - 1}{3}$.

3. Find $(g^{-1} \circ f^{-1})(x)$:

   $$(g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x)) = g^{-1}(x - 1)$$

   $$g^{-1}(x - 1) = \frac{x - 1}{3}$$

Since the results from steps 2 and 3 are the same, it is proven that $(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x)$.

Domain and Range Relationship

The domain of the original function $f$ becomes the range of its inverse function $f^{-1}$, and the range of the original function $f$ becomes the domain of its inverse function $f^{-1}$.