How Do You Find the Inverse of a Function?

To find an inverse function, reverse the original function’s input-output process. At SL, this means reversing values in a table or reading a reflected graph; algebraically deriving an inverse formula is HL-only content.

An inverse function “undoes” the original function. If $f(a)=b$, then $f^{-1}(b)=a$. Applying one function after the other returns the starting value: $f^{-1}(f(x))=x$, wherever the composition is defined.

Therefore, the domain and range exchange roles:

For example, suppose a function contains the ordered pairs $(1,5)$, $(2,7)$, and $(3,9)$. Reverse each pair by swapping its coordinates. Since $f(2)=7$, it follows that . The inverse contains , , and .

Graphically, the graph of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$. Each point ((a,b)) becomes ((b,a)), explaining why domain and range swap.

However, an inverse is itself a function only when the original is one-to-one: every output must come from exactly one input. A many-to-one function fails the horizontal-line test and needs its domain restricted to a one-to-one portion before an inverse function exists.

At HL, for $f(x)=3x-4$, write $y=3x-4$, swap $x$ and to obtain , then solve: . Thus, .

A common misconception is that $f^{-1}(x)=\frac{1}{f(x)}$. This is false: $$ denotes the inverse, while is the reciprocal.

You can verify an answer by composition. For the HL example, $f(f^{-1}(x))=3\left(\frac{x+4}{3}\right)-4=x$, and . This confirms that each function undoes the other on the relevant domain. On a GDC, a graph and its inverse should appear as mirror images across , but algebraic or mapping evidence should still be shown clearly in an examination response.

Exam technique: For SL 2.2, interpret inverse values from mappings, tables, or graphs and exchange domain and range. Finding an inverse formula belongs to AHL 2.7. Always verify that the original function is one-to-one and state any required domain restriction.