Step By Step Calculus » 4.7 - Inverse Functions

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Synopsis
Given a function ff, it is often important to consider the inverse function. Given a function, f: D_f\rightarrow E_ff: D_f\rightarrow E_f, with E_fE_f the range of ff and a function g:D_g\to E_gg:D_g\to E_g with D_g=E_f, \; E_g = D_{f} D_g=E_f, \; E_g = D_{f} we say that they are inverses to one another if
  • f\circ g(y)=yf\circ g(y)=y for y\in E_fy\in E_f, or
  • g\circ f(x)=xg\circ f(x)=x for x\in D_fx\in D_f.
If either one of these holds, the other is automatically valid. When an inverse exists it is unique and written as f^{-1}f^{-1}. Remember that this has nothing to do with the power ff to the -1-1, \displaystyle f^{-1}\ne \frac{1}{f}\displaystyle f^{-1}\ne \frac{1}{f}. It satisfies the cancellation identities:
  • f(f^{-1}(x))=x \quad \forall x \in E_{f}f(f^{-1}(x))=x \quad \forall x \in E_{f}.
  • f^{-1}(f(x))=x \quad \forall x \in D_{f} f^{-1}(f(x))=x \quad \forall x \in D_{f} .
The inverse represents the opposite map to f(x)f(x), allowing recovery of xx if you are given the image y = f(x) y = f(x) under the map ff.
A picture makes the above much more transparent, and also makes it obvious that \left( f^{-1} \right)^{-1} = f . \left( f^{-1} \right)^{-1} = f .

In order for the inverse to exist, a function must be one-to-one. In plain English: it must never take the same value twice. We often replace the cumbersome expression “one-to-one” with the notation 1-11-1. To prove that a given function ff is 1-11-1 we have several alternatives:
  • prove that a\neq ba\neq b implies that f(a)\neq f(b)f(a)\neq f(b). Equivalently, assume that f(a) = f(b)f(a) = f(b) and show that it follows that a = ba = b.
  • apply the horizontal line test, basically just the vertical line test rotated 9090 degrees. This is due to the fact that for the inverse function f^{-1}f^{-1} the independent variable is on the yy axis.
  • reflect the graph of ff across the line y=xy=x, then apply the vertical line test to the reflected graph. This is because, as we shall show, the graph of f^{-1}f^{-1} is the graph of ff reflected in line y = xy = x.
Given a function ff the best way, often the only way, to find a formula for the inverse function is to proceed (and succeed!) algebraically. The procedure is as follows:
Step 0. (Domain) Find the domain of the given function f(x)f(x) .
Step 1. (Set) Set y=f(x)y=f(x).
Step 2. (Solve) Solve for xx in terms of y: \; x=g(y)y: \; x=g(y) If this relation holds for all y\in E_fy\in E_f (the range of ff), then f(x)f(x) is invertible (has an inverse function) and g(y) = f^{-1}(y)g(y) = f^{-1}(y).
Step 3. (Relabel) Relabel g(x) = f^{-1}(x)g(x) = f^{-1}(x).