DWG NO. 1.6 — Lesson 6 of 6

Inverse Functions

Unit 1: Functions and Their Graphs · ~25 min

Objective Find the inverse of a function algebraically and verify that two functions are inverses of each other.

What an inverse does

An inverse function undoes what the original function did. If f takes 3 to 7, then \(f^{-1}\) takes 7 back to 3. Every input/output pair in f gets flipped in \(f^{-1}\) — which is exactly why the graph of \(f^{-1}\) is the graph of f reflected across the line y = x.

Not every function has an inverse that's also a function. For that to work, f has to be one-to-one — no two different inputs can share an output — which you can check with the horizontal line test: if any horizontal line crosses the graph more than once, the function isn't one-to-one, and its inverse won't pass the vertical line test.

Finding an inverse algebraically

Step 1: Replace f(x) with y.
Step 2: Swap x and y.
Step 3: Solve the new equation for y.
Step 4: Replace y with \(f^{-1}\)(x).

Verifying an inverse

Once you have a candidate for \(f^{-1}\), you can check it two ways that must both hold: f(\(f^{-1}\)(x)) = x, and \(f^{-1}\)(f(x)) = x. If plugging one function into the other always returns x, they're genuine inverses.

f(x) = 2x+1 and its inverse \(f^{-1}\)(x) = (x−1)/2 — mirror images across the dashed line y = x

Worked Example 1 · Finding an inverse

ProblemFind the inverse of f(x) = 3x − 4.

1Replace f(x) with y: y = 3x − 4.

2Swap x and y: x = 3y − 4.

3Solve for y: x + 4 = 3y, so y = (x+4)/3.

\(f^{-1}\)(x) = (x+4) / 3

Worked Example 2 · Verifying an inverse

ProblemConfirm f(x) = 2x + 1 and g(x) = (x−1)/2 are inverses by checking f(g(x)).

1f(g(x)) = 2 · [(x−1)/2] + 1 = (x−1) + 1 = x.

2Check the other direction: g(f(x)) = [(2x+1)−1]/2 = 2x/2 = x.

Both compositions return x, so f and g are inverses

Worked Example 3 · Inverse of a rational function

ProblemFind the inverse of f(x) = (x+2) / (x−1), for x ≠ 1.

1Let y = (x+2)/(x−1), then swap: x = (y+2)/(y−1).

2Multiply both sides by (y−1): x(y−1) = y+2, so xy − x = y + 2.

3Collect the y-terms: xy − y = x + 2, then y(x−1) = x+2, so y = (x+2)/(x−1).

\(f^{-1}\)(x) = (x+2) / (x−1) — this function is its own inverse

Guided practice

1.Find the inverse of f(x) = x − 7.
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y = x−7 → swap: x = y−7 → y = x+7. \(f^{-1}\)(x) = x + 7
2.Find the inverse of f(x) = (x/4) + 2.
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y = x/4 + 2 → swap: x = y/4 + 2 → x−2 = y/4 → y = 4(x−2). \(f^{-1}\)(x) = 4x − 8
3.Explain why f(x) = x² does not have an inverse function over its full domain (all real numbers).
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f(x) = x² fails the horizontal line test — for example, f(2) = 4 and f(−2) = 4, so two different inputs share one output. It's not one-to-one, so no inverse function exists over all real numbers (though restricting the domain to x ≥ 0 fixes this).
4.Verify that f(x) = 5x and g(x) = x/5 are inverses by computing f(g(x)).
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f(g(x)) = 5 · (x/5) = x. Confirmed — f(g(x)) = x, so they are inverses.
5.Find the inverse of f(x) = (2x−3) / 5.
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y = (2x−3)/5 → swap: x = (2y−3)/5 → 5x = 2y − 3 → 5x+3 = 2y → y = (5x+3)/2. \(f^{-1}\)(x) = (5x+3) / 2

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