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.
Inverse functions, domain and range
Reverse a one-to-one function and state where the inverse is defined.
Notes
Before you start: rearrange a linear equation and distinguish an input from an output. A domain is the set of permitted inputs; a range is the set of outputs actually attained.
An inverse reverses the input-output mapping. It is a function only when each attained output comes from one input: the original function must be one-to-one on its stated domain. In \(f^{-1}(x)\), the exponent \(-1\) denotes an inverse, not a reciprocal.
Write \(y=f(x)\), solve for \(x\), then exchange the variable names. The domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\). Check both the algebra and these sets.
For a fractional function, retain the excluded denominator value. Rewriting \((ax+b)/(x-c)\) as a constant plus a nonzero multiple of \(1/(x-c)\) can reveal a missing range value.
Recall later
Find the inverse of \(7-2x\) for \(0\le x\le3\). Check: \((7-x)/2\), with inverse domain \([1,7]\).
Common mistakes
Reading inverse notation as 1/f(x).
Giving the old domain to the inverse without finding the original range.
Cancelling or rearranging away an excluded denominator value.
Three worked examples
Example 1
Find the inverse of \(f(x)=4x-7\), for real \(x\).
Show the reasoning
From \(y=4x-7\), \(x=(y+7)/4\). Therefore \(f^{-1}(x)=(x+7)/4\), with domain and range both \(\mathbb R\). Substitution gives \(f(f^{-1}(x))=x\).
Example 2
Find the inverse of \(f(x)=(x+1)/(x-2)\), \(x\ne2\).
Show the reasoning
\(y(x-2)=x+1\), so \(x(y-1)=2y+1\). Thus \(f^{-1}(x)=(2x+1)/(x-1)\), with domain \(x\ne1\). The original form \(1+3/(x-2)\) shows why its range excludes 1.
Example 3
Find the range and inverse domain of \(f(x)=9-2x\), \(1\le x\le5\).
Show the reasoning
The function decreases. Its endpoint outputs are 7 and −1, so its range is \([-1,7]\). The inverse is \((9-x)/2\), defined for \(-1\le x\le7\), with range \([1,5]\).