Inverse Functions

An inverse function \(f^{-1}\) undoes f by swapping its inputs and outputs. It's a genuine function only when f is one-to-one, and its graph is the reflection of f's graph across the line \(y=x\).

By the end you'll be able to find \(f^{-1}(x)\) by swapping x and y and solving, tell when a function has one, and read \(f^{-1}\)'s graph as a mirror image of f across \(y=x\).

Predict: reflect the point (2, 10) across \(y=x\) — where does it land? Drag a to 2 below and turn on "Show inverse" to check.

This plots \(f(x) = 2x + 6\). Drag a to move the point (a, f(a)) along f. Turn on Show inverse to reveal its mirror (f(a), a) on \(f^{-1}\), reflected across the dashed line y = x. Hover either point, or either curve, for exact readouts.

Point (a, f(a)) = (-2.0, 2.0) on f. Turn on "Show inverse" to see its mirror.

f(x) = 2x + 6 reflected across \(y=x\) to get \(f^{-1}(x) = (x-6)/2\)
f(x) = 2x + 6 f⁻¹(x) = (x−6)/2 y = x (mirror line)

An inverse swaps every input/output pair — turning f's graph into \(f^{-1}\)'s graph is just reflecting every point across \(y=x\).

Visual

Swapping the coordinates of a point (a, b) gives (b, a) — and (b, a) is exactly where (a, b) lands when you fold the plane across the diagonal line \(y=x\). Do that to every point on f's graph and you get \(f^{-1}\)'s graph: the mirror image of f across the line y = x, exactly what you're dragging in the diagram above.

Formal

An inverse function \(f^{-1}\) satisfies \(f^{-1}(f(x)) = x\) and \(f(f^{-1}(x)) = x\) — differently, apply f then undo it (or undo it then apply f) and you're back where you started. Because inverting swaps inputs and outputs, the domain of f becomes the range of \(f^{-1}\) and vice versa.

\(f^{-1}\) is itself a function only when f is one-to-one — every output comes from exactly one input. If two inputs shared an output, swapping would force that one output to map back to two different inputs, which a function can't do. Graphically, f is one-to-one exactly when it passes the horizontal line test — no horizontal line crosses it twice.

Applied

A conversion formula and its reverse are inverses of each other: if \(F(c)=\tfrac{9}{5}c+32\) turns Celsius into Fahrenheit, then \(F^{-1}\) turns Fahrenheit back into Celsius. Whenever a rule "does" something to a number, its inverse (if it has one) is the rule that "undoes" it.

Worked example

Find the inverse of \(f(x) = 2x+6\). Write \(y = 2x+6\). Swap x and y: \(x = 2y+6\). Solve for y: \(x-6=2y\), so \(y=(x-6)/2\). So \(f^{-1}(x) = (x-6)/2\). Check: \(f^{-1}(f(x)) = (2x+6-6)/2 = 2x/2=x\) — the x returns, confirming the inverse.

Your turn

Same steps, a different rule: \(f(x) = 3x - 9\). Write \(y = 3x-9\). Swap x and y: \(x = 3y-9\). Solve for y: \(x+9=3y\), so \(y=\) ____. Write the result as \(f^{-1}(x) = \) ____.

Reveal the answer

\(x+9=3y\) gives \(y=(x+9)/3\), so \(f^{-1}(x) = (x+9)/3\). Check by composition: \(f^{-1}(f(x)) = (3x-9+9)/3 = 3x/3 = x\) — confirmed. This f and f⁻¹ are different from the pair in the diagram above, but the swap-and-solve steps are identical.

More info — why f⁻¹ isn't the same as 1/f

\(f^{-1}\) is notation for the inverse function, not the reciprocal — a common mix-up. For \(f(x)=2x+6\), \(1/f(x) = 1/(2x+6)\), which is a completely different function from \(f^{-1}(x)=(x-6)/2\); plug in x = 3 and compare: \(1/f(3) = 1/12\), while \(f^{-1}(3) = -3/2\). And when f is not one-to-one — like \(f(x)=x^2\) over all reals — it has no function inverse at all until you restrict its domain (say \(x\ge0\)), because two inputs like 2 and -2 both map to 4, so undoing 4 wouldn't give a single answer. See the Math is Fun link in Dive deeper for more visuals on the swap.

Check your understanding

Question 1 of 4

Find the inverse of \(f(x) = 5x - 10\).

Question 2 of 4

Which of these explains why \(f(x) = x^2\) does NOT have a function inverse over the entire set of real numbers?

Question 3 of 4

If f and \(f^{-1}\) are genuine inverses of each other, what is \(f(f^{-1}(x))\) for every x in \(f^{-1}\)'s domain?

Question 4 of 4

The graph of a one-to-one function f passes through the point (3, 11). Which point must lie on the graph of \(f^{-1}\)?

Recap

  • \(f^{-1}\) undoes f: \(f^{-1}(f(x))=x\) and \(f(f^{-1}(x))=x\) — swapping inputs and outputs.
  • To find \(f^{-1}\): write \(y=f(x)\), swap x and y, then solve for y.
  • \(f^{-1}\) is a function only when f is one-to-one (passes the horizontal line test); otherwise restrict f's domain first.
  • The graph of \(f^{-1}\) is the reflection of f's graph across the line \(y=x\): a point (a, b) on f becomes (b, a) on \(f^{-1}\).
  • \(f^{-1}\) means "inverse function," never the reciprocal \(1/f\).

Dive deeper

Sources

  • Inverse Functions