Properties of Inverse Function

Property of Composition with Inverse

A function and its inverse undo each other. The order determines the set on which the identity applies: $f\circ f^{-1}$ is the identity on the range of $f$, while $f^{-1}\circ f$ is the identity on the domain of $f$.

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 $f^{-1}(x) = \frac{x - 3}{2}$. Verify the composition property:

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

Both compositions return the original input $x$.

Property of the Inverse of an Inverse

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

Taking an inverse exchanges the input and output of each pair. Exchanging them a second time restores the original function.

Property of the Inverse of a Composition

If we have a composition of two invertible functions with compatible domain and codomain, 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:

Read the composition from right to left: $f^{-1}$ is applied first, then $g^{-1}$.

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 $(f \circ g)(x)$:

   $$\begin{aligned} (f \circ g)(x) &= f(g(x)) \\ &= f(3x) \\ &= 3x + 1 \end{aligned}$$

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

   $$\begin{aligned} (g^{-1} \circ f^{-1})(x) &= g^{-1}(f^{-1}(x)) \\ &= g^{-1}(x - 1) \end{aligned}$$

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

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 $f^{-1}$.