This tutorial explores the concept of a one-to-one (injective) function using algebraic reasoning and graphical methods. Understanding one-to-one functions is essential for studying inverse functions, their properties, and for solving certain types of equations.
Several functions are examined using the horizontal line test, supported by analytical proofs where appropriate. Additional practice problems are available here: Questions on One-to-One Functions .
A function is a rule that assigns to each element of a set called the domain exactly one element in another set called the range.
A function is one-to-one if no two different elements in the domain correspond to the same element in the range.
Algebraically, let \(x_1\) and \(x_2\) be any elements of the domain. A function \(f\) is one-to-one if:
Using the contrapositive, an equivalent statement is:
This form is particularly useful when proving whether a function is one-to-one.
In the Venn diagram below, the function is one-to-one because no two inputs share the same output.
In the next diagram, the function is not one-to-one because the inputs \(-1\) and \(0\) have the same output.
When the graph of a function is known, we can determine whether it is one-to-one using the horizontal line test.
If every horizontal line intersects the graph at most once, the function is one-to-one.
The following graph is not one-to-one because some horizontal lines intersect the graph more than once.
Show algebraically that all linear functions of the form \[ f(x) = ax + b, \quad a \ne 0 \] are one-to-one.
Solution
Assume \(f(x_1) = f(x_2)\). Then:
\[ ax_1 + b = ax_2 + b \] \[ a(x_1 - x_2) = 0 \]Since \(a \ne 0\), it follows that:
\[ x_1 = x_2 \]Therefore, all linear functions with nonzero slope are one-to-one.
Show analytically and graphically that \[ f(x) = -x^2 + 3 \] is not a one-to-one function.
Solution
\[ -x_1^2 + 3 = -x_2^2 + 3 \] \[ -(x_1^2 - x_2^2) = 0 \] \[ -(x_1 - x_2)(x_1 + x_2) = 0 \]This gives:
\[ x_1 = x_2 \quad \text{or} \quad x_1 = -x_2 \]Since equality of outputs does not force \(x_1 = x_2\), the function is not one-to-one.
Show graphically that the following functions are one-to-one:
\[ f(x) = \ln(x), \quad g(x) = e^x, \quad h(x) = x^3 \]Each graph below intersects any horizontal line at most once.
Only one-to-one functions have inverse functions, and their inverses are also one-to-one.
Because logarithmic and exponential functions are one-to-one, we can equate their inputs.
\[ \ln(2x + 3) = \ln(4x - 2) \Rightarrow 2x + 3 = 4x - 2 \] \[ e^{-x + 2} = e^{3x - 8} \Rightarrow -x + 2 = 3x - 8 \]