Definition, horizontal line test, algebraic proofs, and practice — with fully worked solutions
A function \( f(x) \) is one-to-one (injective) if different inputs always produce different outputs. Formally, for any \( A \) and \( B \) in the domain of \( f \):
\[ A \neq B \;\Rightarrow\; f(A) \neq f(B) \]The contrapositive form is far more convenient for algebraic proofs:
\[ f(A) = f(B) \;\Rightarrow\; A = B \]We will use both this contrapositive approach and the horizontal line test to determine whether each function below is one-to-one.
Is the function \( f \) defined by the following set of ordered pairs one-to-one?
\[ f = \{(1,2), (3,4), (5,6), (8,6), (10,-1)\} \]Answer: No.
Two distinct inputs, \( 5 \) and \( 8 \), produce the same output of \( 6 \): \( f(5) = 6 \) and \( f(8) = 6 \). Therefore, \( f \) is not one-to-one.
Is the function \( g \) defined by the following set of ordered pairs one-to-one?
\[ g = \{(-1,2), (0,4), (2,-4), (5,6), (10,0)\} \]Answer: Yes.
Every \( x \)-value in the domain maps to a distinct \( y \)-value (2, 4, −4, 6, 0), so \( g \) is one-to-one.
Is the function \( f(x) = -x^3 + 3x^2 - 2 \) one-to-one?
Answer: No.
The graph below, combined with the horizontal line test, settles the question.
A horizontal line can meet the graph at three different points, meaning three distinct inputs \( x_1, x_2, x_3 \) give the same output. Hence \( f \) is not one-to-one.
Show that every linear function \( f(x) = ax + b \) with \( a \neq 0 \) is one-to-one.
Use the contrapositive. Assume \( f(A) = f(B) \):
\[ aA + b = aB + b \]Subtract \( b \):
\[ aA = aB \]Divide by \( a \neq 0 \):
\[ A = B \]Since \( f(A) = f(B) \) forces \( A = B \), the function is one-to-one.
Show that every function \( f(x) = a(x - h)^2 + k \) with domain \( x \ge h \) (and \( a \neq 0 \)) is one-to-one.
Assume \( f(A) = f(B) \):
\[ a(A - h)^2 + k = a(B - h)^2 + k \]Subtract \( k \) and divide by \( a \):
\[ (A - h)^2 = (B - h)^2 \]Taking square roots gives two possible cases:
\[ A - h = B - h \quad \text{or} \quad A - h = -(B - h) \]The first case gives \( A = B \). For the second case, note that the domain restriction \( x \ge h \) guarantees \( A - h \ge 0 \) and \( B - h \ge 0 \). The equation \( A - h = -(B - h) \) therefore forces both sides to be zero, i.e. \( A - h = 0 \) and \( B - h = 0 \), so again \( A = B \).
In every case, \( f(A) = f(B) \) implies \( A = B \), so \( f \) is one-to-one.
Is \( f(x) = \dfrac{1}{(x - 2)^2} \) one-to-one?
Answer: No.
Evaluate at \( x = 0 \) and \( x = 4 \):
\[ f(0) = \frac{1}{(0 - 2)^2} = \frac{1}{4}, \qquad f(4) = \frac{1}{(4 - 2)^2} = \frac{1}{4} \]Two distinct inputs give the same output, so \( f \) is not one-to-one.
Show that every function \( f(x) = \dfrac{1}{ax + b} \) with \( a \neq 0 \) is one-to-one.
Assume \( f(A) = f(B) \):
\[ \frac{1}{aA + b} = \frac{1}{aB + b} \]Both denominators are nonzero because neither \( A \) nor \( B \) equals \( -\tfrac{b}{a} \) (that value is excluded from the domain). Cross-multiplying gives:
\[ aB + b = aA + b \]Subtract \( b \):
\[ aB = aA \]Divide by \( a \neq 0 \):
\[ B = A \]Hence \( f(A) = f(B) \) implies \( A = B \), so \( f \) is one-to-one.
| Function Type | One-to-One? | Reason |
|---|---|---|
| Linear \( ax + b \), \( a \neq 0 \) | Yes | Dividing by \( a \) gives \( A = B \) |
| Constant | No | Every input gives the same output |
| Quadratic \( x^2 \) on all reals | No | \( f(-1) = f(1) \) |
| Quadratic on \( x \ge h \) | Yes | Monotone on that half-line |
| \( 1/(ax + b) \) | Yes | Reduces to \( A = B \) |
| \( 1/(x - h)^2 \) | No | \( f(h - t) = f(h + t) \) |
| Odd-degree polynomial | Usually, not always | Depends on local extrema (e.g. \( -x^3 + 3x^2 - 2 \) is not 1-1) |
Decide whether each function is one-to-one, then expand the answer below to check your reasoning.
A function \( f \) is one-to-one (injective) if every distinct input produces a distinct output. Equivalently, \( f(A) = f(B) \) implies \( A = B \) for all \( A, B \) in the domain.
A function is one-to-one if and only if every horizontal line intersects its graph at most once. If any horizontal line crosses the graph two or more times, the function is not one-to-one.
No. Because \( f(-1) = f(1) = 1 \), two distinct inputs give the same output. Restricting the domain to \( x \ge 0 \) would make it one-to-one.
Every non-constant linear function \( f(x) = ax + b \) with \( a \neq 0 \) is one-to-one. A constant function (\( a = 0 \)) is not.
The contrapositive form "\( f(A) = f(B) \Rightarrow A = B \)" gives a concrete equation to manipulate algebraically, whereas the direct statement "\( A \neq B \Rightarrow f(A) \neq f(B) \)" is harder to prove directly.
Last updated: October 17, 2026