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Math Practice

Inverse Functions Practice Problems

Fourteen hand-built problems on inverse functions, each fully worked. They start with the one-to-one test that decides whether an inverse exists at all, then drill the swap-and-solve procedure on linear, cubic and rational functions, including the case where the variable appears twice and has to be collected and factored.

  1. 1.

    Find the inverse of \(f(x) = x + 9\). Give \(f^{-1}(x)\) in terms of \(x\).

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    Swap and solve: \(x = y + 9\), so \(y = x - 9\). Adding 9 is undone by subtracting 9.

  2. 2.

    Find the inverse of \(f(x) = 6x\). Give \(f^{-1}(x)\) in terms of \(x\).

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    \(x = 6y\), so \(y = \dfrac{x}{6}\).

  3. 3.

    Is \(f(x) = 7x - 2\) one-to-one? Answer yes or no.

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    If \(7a - 2 = 7b - 2\) then \(a = b\), so yes. Any non-constant line is one-to-one, and no horizontal line crosses it twice.

  4. 4.

    Is \(f(x) = x^2 + 5\) one-to-one? Answer yes or no.

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    \(f(2) = f(-2) = 9\), so two inputs share an output. One counterexample is enough to settle it.

  5. 5.

    If \(f(2) = 11\), what is \(f^{-1}(11)\)?

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    The inverse sends an output back to the input it came from, so \(f^{-1}(11) = 2\).

  6. 6.

    Find the inverse of \(f(x) = 2x - 7\). Give \(f^{-1}(x)\) in terms of \(x\).

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    \(x = 2y - 7\), so \(x + 7 = 2y\) and \(y = \dfrac{x + 7}{2}\).

  7. 7.

    Find the inverse of \(f(x) = 5x - 8\). Give \(f^{-1}(x)\) in terms of \(x\).

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    \(x = 5y - 8\), so \(y = \dfrac{x + 8}{5}\). Check: \(5 \cdot \dfrac{x+8}{5} - 8 = x\).

  8. 8.

    Find the inverse of \(f(x) = \dfrac{x}{3} + 4\). Give \(f^{-1}(x)\) in terms of \(x\).

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    \(x = \dfrac{y}{3} + 4\), so \(x - 4 = \dfrac{y}{3}\) and \(y = 3(x - 4) = 3x - 12\).

  9. 9.

    Find the inverse of \(f(x) = \dfrac{x + 1}{2}\). Give \(f^{-1}(x)\) in terms of \(x\).

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    \(x = \dfrac{y + 1}{2}\), so \(2x = y + 1\) and \(y = 2x - 1\).

  10. 10.

    Find the inverse of \(f(x) = x^3 + 1\). Give \(f^{-1}(x)\) in terms of \(x\).

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    \(x = y^3 + 1\), so \(y^3 = x - 1\) and \(y = \sqrt[3]{x - 1}\). A cubic is one-to-one, so this inverse is a function everywhere.

  11. 11.

    \(f\) has domain \([1,\ 5]\) and range \([0,\ 8]\). What is the domain of \(f^{-1}\)? Give it in interval notation.

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    Domain and range swap, so the domain of \(f^{-1}\) is the range of \(f\), which is \([0,\ 8]\).

  12. 12.

    Find the inverse of \(f(x) = \dfrac{1}{x}\). Give \(f^{-1}(x)\) in terms of \(x\).

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    \(x = \dfrac{1}{y}\) gives \(y = \dfrac{1}{x}\). This function is its own inverse — reflecting its graph in \(y = x\) leaves it unchanged.

  13. 13.

    Find the inverse of \(f(x) = \dfrac{3x}{x + 2}\). Give \(f^{-1}(x)\) in terms of \(x\).

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    \(x = \dfrac{3y}{y + 2}\) gives \(x(y + 2) = 3y\), so \(xy + 2x = 3y\). Collect the \(y\)-terms: \(2x = 3y - xy = y(3 - x)\), so \(y = \dfrac{2x}{3 - x}\).

  14. 14.

    Celsius to Fahrenheit is \(F(c) = \dfrac{9}{5}c + 32\). Give the inverse \(F^{-1}(x)\) in terms of \(x\).

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    \(x = \dfrac{9}{5}y + 32\) gives \(x - 32 = \dfrac{9}{5}y\), so \(y = \dfrac{5}{9}(x - 32)\) — the familiar Fahrenheit-to-Celsius formula.
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Inverse Functions Practice Problems

Two lessons feed this set: One-to-One Functions, which decides whether an inverse exists at all, and Inverse Functions, which finds it. They belong in one practice set because the first question is always “does this even have an inverse?”

How to use this set

The procedure never changes: write , swap and , solve for . Doing it in that order is what stops the intermediate lines becoming meaningless.

There is a built-in check on every answer. The inverse should undo the original’s operations in reverse order — if multiplies by 3 then adds 6, then subtracts 6 then divides by 3. If your answer does not read that way backwards, something has gone wrong.

Type the inverse as an expression in : (x + 7)/2. For a cube root, nthRoot(x - 1, 3) or \sqrt[3]{x-1} are both accepted.

Common mistakes to watch for

Reading as a reciprocal. It is the inverse function.

Solving before swapping. Swap and first.

Forgetting the domain restriction. The inverse’s domain is the original’s range.

Inverting something that is not one-to-one. Restrict the domain first, or the “inverse” is not a function.

Stopping when the variable appears twice. Collect and factor — that is the whole technique in problem 13.

Where to go next

Function Operations and Composition is where the verification step comes from, and Domain and Range drills the two sets that swap places.

Frequently Asked Questions

What is an inverse function?+

One that undoes the original. If takes 3 to 11, then takes 11 back to 3. Formally and .

How do you find an inverse?+

Write , swap and , then solve the new equation for . Swapping first and solving second is what keeps the method reliable.

Does f⁻¹ mean 1 over f?+

No. is the inverse function; is the reciprocal. The is a label, not an exponent, and the two are almost never equal.

Which functions have inverses?+

Only one-to-one ones. If two inputs share an output, reversing would give that value two outputs, which no function may do. A function that fails can often be inverted on a restricted domain.

What happens to the domain and range?+

They swap. The domain of is the range of , and vice versa. This is often the only way to get the inverse's domain right, since the formula alone may accept more than it should.

What do I do when the variable appears twice?+

Multiply out to clear the fraction, gather every term containing the variable on one side, factor it out, then divide. Problem 13 is the standard example.

More practice problems

Function Operations and Composition Practice Problems Function Notation Practice Problems Domain and Range Practice Problems