2.5 The Inverse Function: $f^{-1}$

1. Formal Definition

The inverse procedure essentially reverses a given mapping. If $f: \mathbb{R} \rightarrow \mathbb{R}$, its inverse function $f^{-1}$ satisfies:

$f(x) = y \Leftrightarrow f^{-1}(y) = x$
  • The domain of $f$ becomes the range of $f^{-1}$, and vice-versa: $D_{f^{-1}} = R_f$ and $R_{f^{-1}} = D_f$.
  • The composition of inverses establishes the identity function: $(f^{-1} \circ f)(x) = x$ and $(f \circ f^{-1})(x) = x$.
  • $(f^{-1})^{-1} = f$.

2. Determining the Inverse Function

A systematic three-step procedure evaluates the algebraic inverse formula:

  1. Set the function equal to y: $f(x) = y$.
  2. Solve algebraically for x.
  3. Keep the solution, but interchange the variables (replace $y$ with $x$).

EXAMPLE 1

Let $f(x) = 3x + 5$. Find (a) $f^{-1}(x)$ and (b) $f^{-1}(11)$.

(a) Set $3x + 5 = y$. Solve for $x$: $3x = y - 5 \Rightarrow x = \frac{y - 5}{3}$. Swap variables: $\mathbf{f^{-1}(x) = \frac{x - 5}{3}}$.
(b) Substitute into inverse: $f^{-1}(11) = \frac{11 - 5}{3} = \mathbf{2}$.
Alternative logical route: $f^{-1}(11) = x \Rightarrow f(x) = 11 \Rightarrow 3x + 5 = 11 \Rightarrow x = \mathbf{2}$.

EXAMPLE 2

Let $f(x) = 2x^2 - 1$ where $x \ge 0$. Find $f^{-1}(x)$.

Set $2x^2 - 1 = y \Rightarrow 2x^2 = y + 1 \Rightarrow x^2 = \frac{y + 1}{2}$.
Because domain restricts $x \ge 0$, select positive root: $x = \sqrt{\frac{y+1}{2}}$.
Result: $\mathbf{f^{-1}(x) = \sqrt{\frac{x+1}{2}}}$.

EXAMPLE 3

Let $f(x) = \frac{x+1}{x+2}$. Find $f^{-1}(x)$.

Set $\frac{x+1}{x+2} = y \Rightarrow x + 1 = y(x+2) \Rightarrow x + 1 = yx + 2y$.
Factor x terms: $x - yx = 2y - 1 \Rightarrow x(1-y) = 2y - 1 \Rightarrow x = \frac{2y-1}{1-y}$.
Result: $\mathbf{f^{-1}(x) = \frac{2x-1}{1-x}}$.

EXAMPLE 4 (Composite Inverses)

Let $f(x) = 1-2x$ and $g(x) = 1/x$.

Inverse property to note: The inverse of a composition reverses the function order: $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$.

3. Graphical and Analytical Properties

Graphing: The graph of $f^{-1}$ is a strict geometric reflection of $f$ across the diagonal line $y=x$. Intersection points between an increasing function and its inverse can be found by evaluating the simpler equation $f(x) = x$.
Self-Inverse Functions: Functions that satisfy $f^{-1} = f$ are inherently symmetric across $y=x$. Examples include $f(x) = 1/x$ and certain rational functions.
Presupposition (HL): For an inverse function to exist over a domain, different inputs must strictly map to different outputs. Therefore, a function must be One-to-One (passing the horizontal line test) to possess an inverse.