Identifying one-to-one functions

A function is one-to-one when every element of the range of a function corresponds to exactly one element of the domain. That may not make much sense on first reading, so we will try to get to the concept underpinning the definition.

Consider the function $f\left(x\right)=2x+3$f(x)=2x+3 with the restricted domain given as $\left\{-3,0,3,5\right\}${−3,0,3,5}. The four function values that make up the range become $f\left(-3\right)=-3$f(−3)=−3, $f\left(0\right)=3$f(0)=3, $f\left(3\right)=9$f(3)=9 and $f\left(5\right)=13$f(5)=13.

Thus each element of the domain $\left\{-3,0,3,5\right\}${−3,0,3,5} is mapped, one-to-one, onto the four elements of the range $\left\{-3,3,9,13\right\}${−3,3,9,13}. The four different elements of the domain are mapped to four different elements of the range.

Now consider the function $f\left(x\right)=x^2+3$f(x)=x2+3 with the same restricted domain $\left\{-3,0,3,5\right\}${−3,0,3,5}. The new function values make up the range given as $\left\{12,3,12,28\right\}${12,3,12,28}.

With this new function note that the two elements $-3$−3 and $3$3 are mapped to the same element $12$12 in the range. This new function is thus not one-to-one. In fact we call such a function many-to-one.

If we again take the function $f\left(x\right)=2x+3$f(x)=2x+3 but this time consider the domain as the natural domain - that is, the complete set of real numbers.

We can be assured that the function is still one-to-one by simply thinking about the graph of $y=2x+3$y=2x+3. It is in a constant state of rising and therefore the real values of x are being mapped to distinct real function values.

We can also be assured that the function $f\left(x\right)=x^2+3$f(x)=x2+3, with a natural domain of real numbers, is not one-to-one because the graph falls to a minimum turning point of $\left(0,3\right)$(0,3) and then begins rising again. Thus there are, for all points other than that minimum turning point, two real values in the domain being mapped to each real element in the range.

The first three functions, $G1$G1 given by $f\left(x\right)=x^2$f(x)=x2, $G2$G2 given by $f(x)=\left|x\right|$f(x)=|x| and the semicircle $G3$G3 given by $f\left(x\right)=\sqrt{4-x^2}$f(x)=√4−x2 are not one-to-one because an imaginary ruler can be placed horizontally onto all three graphs in such a way as to cut them more than once. The graph $G4$G4 however, representing the cube root of $x$x is a one-to-one function.

Practice questions

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QUESTION 3