Finding the Inverse of a Function

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Idea

To reverse a function, solve its output equation for the original input. First establish one-to-one behavior on the chosen domain; otherwise reversing the relation may give multiple outputs.

Visual guide

VISUAL GUIDESwap the coordinates to invert

f(x) = 2x + 1
f⁻¹(x) = (x − 1)/2
Reflection line y = x

For f(x) = 2x + 1, the inverse is (x − 1)/2. Points (0, 1) and (1, 0) exchange coordinates across y = x. The inverse reverses the input/output relationship; it is not the reciprocal 1/f(x).

Method

Write $y=f(x)$ , solve for $x$ in terms of $y$ , then rename the input of the inverse. Domain and range exchange roles. Verify by composition on the appropriate domains.

Worked example

For $f(x)=3x-5$ , solve $y=3x-5$ to get $x=(y+5)/3$ . Hence $f^{-1}(x)=(x+5)/3$ . For $g(x)=x^2$ on $x\ge0$ , take the nonnegative branch: $g^{-1}(x)=\sqrt x$ , also defined for $x\ge0$ .

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Find the inverse of f(x)=3x−2 on ℝ.

f⁻¹(x)=1/(3x−2)

f⁻¹(x)=(x+2)/3

f⁻¹(x)=(x−2)/3

Hint 1 · Find a starting point

Undo subtraction before undoing multiplication.

Hint 2 · Take the next step

Solve y=3x−2 for x.

Show the reasoning

Answer: f⁻¹(x)=(x+2)/3

x=(y+2)/3; renaming the inverse input gives (x+2)/3.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Common mistake

Writing both $\pm\sqrt x$ does not define a single inverse function; the original domain decides which branch to use.

Check your understanding

Find the inverse of $f(x)=\sqrt{x-1}$ on $[1,\infty)$ .

Show answer

$f^{-1}(x)=x^2+1$ on $[0,\infty)$ .

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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