Learn the method, study an example, then explain your reasoning in an exam-style problem. Work on paper before opening the solution and compare each step with the marking guidance.
A function and its inverse on a graph
Reflect coordinates and interpret common points without assuming a false general rule.
Notes
Before you start: plot coordinates and solve simultaneous equations.
If \((a,b)\) lies on \(y=f(x)\), then \((b,a)\) lies on \(y=f^{-1}(x)\). The graphs are reflections in \(y=x\). Swap vertices and endpoints as well as ordinary points; transfer domain and range restrictions.
A fixed point solves \(f(x)=x\) and lies on both graphs. For a strictly increasing real function, every common point must be on \(y=x\): if \(f(x)=y\) and \(f(y)=x\), then \(x
Do not use that conclusion for every invertible function. A decreasing function can have common points off \(y=x\); for example \(f(x)=6-x\) is its own inverse, so the two entire graphs coincide.
Recall later
Reflect \((-2,5)\) in \(y=x\): the image is \((5,-2)\). The signs stay attached to their coordinates.
Common mistakes
Reflecting in an axis rather than y=x.
Swapping the coordinates but dropping a branch restriction.
Assuming every intersection with an inverse is a fixed point, even for decreasing functions.
Three worked examples
Example 1
The graph of \(f(x)=2x+2\) contains \((1,4)\) and \((3,8)\). Locate the reflected points.
Show the reasoning
The inverse graph contains \((4,1)\) and \((8,3)\). Its equation is \(y=(x-2)/2\). The common fixed point solves \(2x+2=x\), giving \((-2,-2)\).
Example 2
Reflect the branch \(y=(x-2)^2+1\), \(x\ge2\).
Show the reasoning
Its vertex \((2,1)\) becomes \((1,2)\); its point \((3,2)\) becomes \((2,3)\). The inverse is \(y=2+\sqrt{x-1}\), \(x\ge1\), with \(y\ge2\).
Example 3
Find the inverse of \(f(x)=6-x\) and describe the two graphs.
Show the reasoning
Rearranging \(y=6-x\) gives \(x=6-y\), so \(f^{-1}(x)=6-x\). The graphs coincide. For example \((1,5)\) is common even though it is not on \(y=x\).