Why Inverses Exchange Inputs and Outputs
A function takes an input and assigns it an output. An inverse function reverses that assignment: it takes the original output and returns the original input. This reversal is possible as a function only when each output came from exactly one input.
For example, if \(f(2)=7\), an inverse function must satisfy \(f^{-1}(7)=2\). The input and output have traded places. This simple idea determines both how to find an inverse and how to interpret its graph.
The inverse’s domain is \(f(A)\), the range of the original function, and its range is \(A\), the original domain. If a function is specified with a larger codomain than its range, its inverse is still defined on the range of actual outputs, not on values the function never produces.
The notation \(f^{-1}\) means the inverse function; it does not mean the reciprocal \(1/f\). These are different operations. In particular, the reciprocal of \(f(x)\) is written \(1/f(x)\), when defined, while \(f^{-1}(x)\) undoes the action of \(f\).
Finding an Inverse by Swapping Coordinates
Suppose the point \((a,b)\) lies on the graph of \(f\). By the definition of a function’s graph, \(b=f(a)\). The inverse reverses this assignment, so \(f^{-1}(b)=a\); therefore \((b,a)\) lies on the graph of \(f^{-1}\). To find an inverse algebraically, write the original input-output relationship, exchange the variables, and solve for the new output.
Worked Example: Inverting a Linear Function
Find the inverse of \(f(x)=3x-5\), and identify a pair of corresponding graph points.
Begin with the input-output equation and exchange the variables:
Solve the exchanged equation for \(y\). Add \(5\) to both sides, then divide by \(3\):
Therefore \(f^{-1}(x)=\frac{x+5}{3}\). The original function is one-to-one because if \(3x_1-5=3x_2-5\), then \(3x_1=3x_2\), so \(x_1=x_2\).
Check the inverse by composing in both orders:
For a point check, \(f(2)=3(2)-5=1\), so \((2,1)\) lies on the graph of \(f\). The inverse gives \(f^{-1}(1)=\frac{1+5}{3}=2\), so \((1,2)\) lies on the inverse graph. The coordinates have exchanged places.
Why the Graphs Reflect Across \(y=x\)
The coordinate swap gives a geometric interpretation. Reflecting a point across the line \(y=x\) exchanges its horizontal and vertical coordinates: \((a,b)\) becomes \((b,a)\). Since every point on the graph of \(f\) corresponds to a swapped point on the graph of \(f^{-1}\), the two graphs are reflections of one another across \(y=x\).
This reflection statement follows directly from the point correspondence. If \((a,b)\) is on the graph of \(f\), then \(b=f(a)\). By the definition of the inverse, \(f^{-1}(b)=a\), so the reflected point \((b,a)\) is on the inverse graph. Reversing the argument shows that every point on the inverse graph reflects back to a point on the original graph. Thus the correspondence works in both directions.
The line \(y=x\) is the axis of this reflection. A point on that line has equal coordinates, so it reflects to itself. Where the graphs intersect the line \(y=x\), their coordinates satisfy \(f(x)=x\). This observation describes possible intersection points; it does not say that every function’s graph intersects the line.
When a Domain Restriction Is Necessary
A function has an inverse function only if it is one-to-one on its domain. If two different inputs produce the same output, reversing the assignment would require that one output be sent to two inputs. That would not be a function.
On a graph, a function is one-to-one precisely when every horizontal line intersects its graph at most once. If some horizontal line crosses the graph twice, the two distinct input values have the same output, so the function fails the one-to-one condition. A suitable restriction of the original domain may remove this problem.
Worked Example: Restricting a Squared Function
The function \(f(x)=x^2\), with domain \(\mathbb{R}\), is not one-to-one: \(f(2)=4\) and \(f(-2)=4\), although \(2\ne-2\). Restrict its domain to \([0,\infty)\), then find and check its inverse.
On the restricted domain, write the function and its range:
This restricted function is one-to-one. If \(x_1,x_2\ge 0\) and \(x_1^2=x_2^2\), then \((x_1-x_2)(x_1+x_2)=0\). Since \(x_1+x_2\ge0\), equality of the squares for nonnegative inputs implies \(x_1=x_2\).
Exchange input and output in \(y=x^2\), then solve for \(y\):
The nonnegative square root is required because the inverse’s outputs must lie in the restricted original domain \([0,\infty)\). Thus \(f^{-1}(x)=\sqrt{x}\), with domain \([0,\infty)\).
Check both compositions, respecting their domains:
For a point check, \(f(3)=3^2=9\), so \((3,9)\) is on the restricted graph. The inverse gives \(f^{-1}(9)=\sqrt{9}=3\), so \((9,3)\) is on the inverse graph.
Choosing and Checking a Restricted Branch
A domain restriction must do two things: make the function one-to-one and preserve the part of the function you intend to use. For a parabola, either side of its vertex can provide a one-to-one domain, but the resulting inverse depends on which side is chosen. The range of the restricted function becomes the inverse’s domain.
Worked Example: Inverting a Shifted Parabola on One Branch
Let \(f(x)=(x-2)^2-3\) have domain \([2,\infty)\). Find its inverse, state the inverse’s domain and range, and verify the result.
For \(x\ge2\), the function increases from its value at \(x=2\), which is \(-3\), without turning back. Its range is therefore \([-3,\infty)\). Exchange input and output:
Add \(3\) to both sides and take square roots:
The original domain is \([2,\infty)\), so inverse outputs must satisfy \(y\ge2\). Choose the positive square root:
Verify both compositions. For \(x\ge2\):
The last equality holds because \(x\ge2\), so \(|x-2|=x-2\). For \(x\ge-3\):
As a point check, \(f(4)=(4-2)^2-3=4-3=1\). Hence \((4,1)\) lies on the original graph, and \(f^{-1}(1)=2+\sqrt{1+3}=2+2=4\), giving the swapped point \((1,4)\).
A Common Mistake: Keeping Both Square-Root Branches
A frequent error is to solve an equation for \(y\) and treat both solutions as one inverse function, even though the original function has not been restricted to a one-to-one domain.
For example, starting from \(y=x^2\), swapping variables gives \(x=y^2\), so algebraically \(y=\pm\sqrt{x}\). But for \(x=4\), this relation gives \(y=2\) and \(y=-2\). It assigns two outputs to the same input, so it is not a function. The original \(x^2\) function on all real inputs is not one-to-one, and it cannot have an inverse function on that full domain.
The correction is to choose a domain restriction first. On \([0,\infty)\), the inverse is \(y=\sqrt{x}\); on \((-\infty,0]\), the inverse is \(y=-\sqrt{x}\). Each choice gives a single output for every input in the corresponding range. The correct branch is determined by the chosen original domain, not by a preference made after solving.
Check Your Understanding
Answer each question using the ideas of coordinate exchange, one-to-one functions, and domain restriction.
- If \(f(5)=2\), what point must lie on the graph of \(f^{-1}\)?
- Why does a horizontal line that intersects a function’s graph twice show that the function is not one-to-one?
- Find the inverse of \(f(x)=x+4\) by exchanging input and output, and state the inverse’s domain.
- Why does \(f(x)=x^2\) on all real numbers fail to have an inverse function?
- If \(f(x)=x^2\) is restricted to \((-\infty,0]\), which square-root branch gives the inverse, and why?
