Inverse Functions

Undoing a function to find the one that reverses it.

A GPS calculates your position from a signal travel time. The device converts time into distance. But engineers who built the system had to first answer the reverse question: given a distance, what travel time produced it? They needed the function that undoes the original. That reversal has a name: the inverse function.

A function takes an input and produces an output. The inverse function takes that output and hands back the original input. If $f(3) = 7$ , then the inverse, written $f^$ , satisfies KaTeX can only parse string typed expression. The notation $f^$ does not mean $\frac$ — it is a name for the reverse function, not a reciprocal.

The core idea is this: a function and its inverse undo each other completely. Feed the output of $f$ into $f^$ and you get back where you started.

f^,(f(x)) = x \quad \text, \quad f(f^,(x)) = x

To find the inverse of a function algebraically, swap $x$ and $y$ , then solve for $y$ . The reason swapping works is that the input and output trade roles — what was the output becomes the new input, and what was the input becomes the new output.

Finding the inverse of f(x) = 2x + 6

y = 2x + 6

Write f(x) as y — this makes the swap cleaner.

x = 2y + 6

Swap x and y. This is the core move — the output is now the input.

x - 6 = 2y

Subtract 6 from both sides to isolate the y term.

y = \frac{x - 6}{2}

Divide both sides by 2 to solve for y.

f^{-1}(x) = \frac{x - 6}{2}

Rename y as the inverse function.

Now verify that these two functions actually undo each other. Plug $f(x)$ into $f^$ and check that you get $x$ back.

Verifying the inverse of f(x) = 2x + 6

f^{-1}(f(x)) = \frac{(2x + 6) - 6}{2}

Replace the input of f⁻¹ with the full expression for f(x).

= \frac{2x}{2}

The 6 and -6 cancel.

= x \checkmark

You get x back. The functions undo each other — the inverse is correct.

There is also a graphical side to this. Every point $(a, b)$ on the graph of $f$ becomes the point $(b, a)$ on the graph of $f^$ . Swapping the coordinates of every point is the same as reflecting the graph across the line $y = x$ . The line $y = x$ acts as a mirror between a function and its inverse.

Interactive graph — scroll to zoom, drag to pan

The red line is $f(x) = 2x + 6$ . The blue line is KaTeX can only parse string typed expression. The green line is $y = x$ . Notice the two function graphs are exact mirror images across the green line.

One more thing to track: the domain and range swap between a function and its inverse. If $f$ has domain $A$ and range $B$ , then $f^$ has domain $B$ and range $A$ . This matters especially when the original function has a restricted domain.

Here is a harder example where the domain restriction plays a role.

Finding the inverse of f(x) = x^{2} + 1, restricted to x ≥ 0

y = x^2 + 1

Write as y.

x = y^2 + 1

Swap x and y.

x - 1 = y^2

Subtract 1 from both sides.

y = \sqrt{x - 1}

Take the square root. You take the positive root only because the original domain was x ≥ 0, which means the output of the inverse must also be non-negative.

f^{-1}(x) = \sqrt{x - 1}, \quad x \geq 1

The domain of the inverse is x ≥ 1 because the range of f(x) = x² + 1 on x ≥ 0 starts at 1.

Without the restriction $x \geq 0$ on the original function, the parabola fails the horizontal line test — a single output value corresponds to two different input values, so no true inverse exists. Restricting the domain forces the function to be one-to-one, which is the requirement for an inverse to exist.

Practice Questions

f(x) = 3x - 9

f(x) = \frac{x + 4}{5}

f(x) = x^2 - 3, \quad x \geq 0

Regents Corner

On Part II and Part III of the Algebra II Regents, inverse function problems often ask you to find the inverse algebraically and then verify it. Showing both KaTeX can only parse string typed expression and KaTeX can only parse string typed expression earns full verification credit. Showing only one composition is partial credit on a two-point verification question.

Students see f⁻¹(x) and write it as 1/f(x). These are completely different things. If f(x) = 2x + 6, then 1/f(x) = 1/(2x+6), which is a reciprocal. The inverse f⁻¹(x) = (x−6)/2, which undoes the original function. The superscript −1 on a function name means inverse, not reciprocal.

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