Section 6.9 One-to-one Functions
Not all functions have inverse functions. Here’s an example.
Example 6.43.
Let \(f(x) = x^2 + 2\text{.}\) I evaluated at a real number \(x\) and found \(f(x) = 6\text{.}\) Can you determine my value of \(x\text{?}\)
We need to solve the equation \(f(x) = 6\) for \(x\text{.}\) This will prove to be a problem.
\begin{align*}
f(x) &= 6\\
x^2+2 &= 6\\
x^2 &= 4\\
x &= \pm \sqrt{4} = \pm 2
\end{align*}
Thus, \(x=2\) or \(x = -2\) and it is impossible to determine which value I evaluated \(f\) at to get \(6\text{.}\) So while we have limited the possibilities to two different possible values, we cannot explicitly determine the input to the function from the output.
The problem here is two distinct inputs
\(-2 \neq 2\) result in the same output
\(f(-2)=f(2)\text{.}\) Graphically, the horizontal line
\(y=6\) intersects the graph at two different points corresponding to the two distinct inputs.
More generally, if \(f(x) = y\text{,}\) then solving for \(x\) gives
\begin{equation*}
x = \pm \sqrt{y- 4}
\end{equation*}
and \(x\) is not a function of \(y\text{.}\) Therefore, there is no inverse function for \(f\text{.}\)
To avoid this situation, we require that each output of a function \(f\) corresponds to exactly one input. Such functions are called one-to-one functions.
Definition 6.44. One-to-one Functions.
A function \(f\) is one-to-one if whenever \(a\neq b\) are in the domain of \(f\text{,}\) then \(f(a)\neq f(b)\text{.}\) That is, distinct inputs to \(f\) result in distinct outputs.
Example 6.45.
Let \(f(x) = x^2+2\text{.}\) Since \(f(-2)=f(2)\text{,}\) \(f\) is not one-to-one.
Graphically, if distinct inputs are to give distinct outputs, a horizontal line should never pass through the graph more than once. This is called the horizontal line test.
Principle 6.46. Horizontal Line Test.
A function is one-to-one if and only if no horizontal line intersects the graph of \(y=f(x)\) more than once.
Example 6.47.
Let
\(f(x) = x^3+2\text{.}\) The graph passes the Horizontal Line Test.
Therefore
\(f\) is one-to-one.
To conclude, one-to-one functions have inverse functions.
Theorem 6.48. One-to-one Functions have Inverse Functions.
A function \(f\) has an inverse function if and only if \(f\) is one-to-one.
Example 6.49.
Let \(g(x) = \frac{2x-1}{3-x}\text{.}\) Given that \(g\text{,}\) find \(g^{-1}(x)\text{.}\)
Solution.
We need to solve \(y=g(x)\) for \(x\text{.}\)
\begin{align*}
y &= \frac{2x-1}{3-x}\\
y(3-x) &= 2x-1\\
3y-xy &= 2x-1\\
3y+1 &= 2x-xy\\
3y+1 &= x(2-y)\\
\frac{3y+1}{2-y} &= x
\end{align*}
So that
\begin{equation*}
g^{-1}(y) = \frac{3y+1}{2-y}.
\end{equation*}
If you happen to prefer \(x\) as the input, then write
\begin{equation*}
g^{-1}(x) = \frac{3x+1}{2-x}.
\end{equation*}
