MISSION · Algebra
Inverse and Composite Functions
Doing one thing after another, and getting back to where you started
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LEVEL 2 · LEARN
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The idea in one sentence
Composition feeds one function’s output into another; an inverse is the function that takes you back.
Composition
\[(f \circ g)(x) = f(g(x))\]Do \(g\) first, then \(f\). The inner function goes first, which looks backwards because it’s written second — a genuine source of confusion, and worth reading carefully every time.
With \(f(x) = x^2\) and \(g(x) = x + 3\):
\[f(g(2)) = f(5) = 25 \qquad g(f(2)) = g(4) = 7\]Order matters. Squaring after adding is not the same as adding after squaring, and the two compositions are different functions, not two routes to one answer.
Think of it as two machines bolted together. What comes out of the first is what goes into the second, so the first machine’s outputs have to be acceptable inputs for the second — a restriction that occasionally shrinks the domain of the composition.
Inverses undo
\[f^{-1}(f(x)) = x\]If \(f\) doubles and adds 1, then \(f^{-1}\) subtracts 1 and halves. Notice that the operations are reversed and taken in the opposite order — same as taking off your socks and shoes. Last on, first off.
To find an inverse algebraically:
- Write \(y = f(x)\).
- Swap \(x\) and \(y\).
- Solve for \(y\).
For \(f(x) = 2x + 1\): swapping gives \(x = 2y + 1\), and solving gives \(y = \frac{x-1}{2}\). That’s \(f^{-1}\).
Not every function has one
An inverse has to be a function itself, which means one input, one output. So \(f\) must never send two different inputs to the same place — reversing that would demand two outputs from one input.
\(f(x) = x^2\) fails: both 3 and \(-3\) give 9, so \(f^{-1}(9)\) would have to be both.
The fix is to restrict the domain. Allow only \(x \ge 0\) and squaring becomes reversible, with \(\sqrt{x}\) as its inverse. That’s why the square root symbol means the positive root only — the restriction is built into the definition.
The test on a graph is the horizontal line test: if any horizontal line meets the graph twice, there’s no inverse over that domain. It’s the vertical line test turned on its side, which makes sense, because an inverse’s graph is the original reflected.
The reflection
The graph of \(f^{-1}\) is the graph of \(f\) reflected in the line \(y = x\).
Swapping \(x\) and \(y\) in the algebra is exactly swapping the axes in the picture, so every point \((a, b)\) on \(f\) becomes \((b, a)\) on \(f^{-1}\). That also explains why the exponential and logarithm graphs are mirror images: they’re inverse functions, and \(\log_b(b^x) = x\) says so.
Common mix-ups
- Reading \(f^{-1}\) as \(\frac{1}{f}\). It’s the inverse function, not a reciprocal.
- Applying the outer function first in a composition. Inner first.
- Inverting the operations without reversing the order.
- Assuming every function has an inverse. It needs to be one-to-one.
Check yourself
For \(f(x) = 3x - 6\), find \(f^{-1}\) and check that \(f^{-1}(f(4)) = 4\).
Swapping and solving gives \(f^{-1}(x) = \frac{x+6}{3}\). And \(f(4) = 6\), then \(f^{-1}(6) = \frac{12}{3} = 4\). Back where we started.
LEVEL 3 · REFLECT
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