Twelfth Grade · Math · week 4

Inverse functions

Find the inverse of a function algebraically, verify it by composition, and restrict a domain to make a function invertible.

A function's inverse, written f^-1(x), undoes exactly what f did: if f turns 5 into 13, then f^-1 turns 13 back into 5. To find an inverse algebraically, swap x and y in the equation y=f(x), then solve for the new y -- that solved equation is f^-1(x).

Not every function has an inverse that is ALSO a function. f must be one-to-one (each output comes from only one input -- check this on a graph with the horizontal line test) for f^-1 to be a genuine function. A function like f(x)=x^2 fails this test over its whole domain, but restricting the domain to x>=0 makes it one-to-one, and THEN it has an inverse: sqrt(x).

Always verify a candidate inverse with composition: f(f^-1(x)) should simplify to x, and f^-1(f(x)) should too. If either check fails, something was solved incorrectly.

📋 For the teacher
Teach it (10–15 min):
  • Say the definition concretely before any algebra: the inverse undoes f -- whatever f did to a number, f^-1 does the opposite, in the opposite order.
  • Make "swap x and y, then solve for y" a fixed, three-word ritual (swap, then solve) so the method never feels mysterious.
  • Always run the composition check on at least one example together -- trusting the algebra without verifying is how sign errors slip through unnoticed.
  • Use the horizontal line test on a quick sketch to explain WHY x^2 needs a restricted domain before it can have an inverse -- the picture makes the one-to-one requirement obvious.
Talk about it:
  • Why does swapping x and y in the equation capture the idea of "undoing" a function?
  • What does it mean for a function to be one-to-one, and how does the horizontal line test check for it?
  • If f and f^-1 are both graphed on the same axes, what is special about how the two graphs relate to each other?

Mastery looks like: They find an inverse algebraically by swapping and solving, verify it with composition, and can explain why restricting a domain can make an otherwise non-invertible function invertible.

Common stumbles: Solving for x instead of y after swapping; forgetting that a restricted-domain inverse is only valid alongside that same restriction.

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