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Function Operations and Composition Practice Problems

Fifteen hand-built problems covering both ways of building a new function from two old ones: arithmetic combination and composition. They drill the sum, difference, product and quotient at a point and as a formula, the extra domain restriction on a quotient, evaluating f(g(x)) inside out, why the order matters, and decomposing a function back into two pieces.

  1. 1.

    For \(f(x) = x + 2\) and \(g(x) = x^2\), find \((f + g)(3)\).

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    \(f(3) = 5\) and \(g(3) = 9\), so \((f+g)(3) = 5 + 9 = 14\).

  2. 2.

    For \(f(x) = x + 2\) and \(g(x) = x^2\), find \((f - g)(3)\).

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    \(5 - 9 = -4\). Order matters here — \((g - f)(3)\) would be \(+4\).

  3. 3.

    For \(f(x) = x + 2\) and \(g(x) = x^2\), find \((fg)(3)\).

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    \(5 \times 9 = 45\).

  4. 4.

    For \(f(x) = x + 4\) and \(g(x) = 2x\), find \(f(g(3))\).

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    Inner first: \(g(3) = 6\). Then \(f(6) = 6 + 4 = 10\).

  5. 5.

    For \(f(x) = x + 4\) and \(g(x) = 2x\), find \(g(f(3))\).

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    \(f(3) = 7\), then \(g(7) = 14\). Compare with the previous problem — swapping the order changed the answer.

  6. 6.

    For \(f(x) = x + 2\) and \(g(x) = x^2\), write \((f + g)(x)\) as a formula in \(x\).

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    \((x + 2) + x^2 = x^2 + x + 2\).

  7. 7.

    For \(f(x) = 3x^2 + 2\) and \(g(x) = x^2 - 4x + 1\), write \((f - g)(x)\) as a formula in \(x\).

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    \((3x^2 + 2) - (x^2 - 4x + 1) = 3x^2 + 2 - x^2 + 4x - 1 = 2x^2 + 4x + 1\). Every sign inside \(g\) flips — keep the bracket while subtracting.

  8. 8.

    For \(f(x) = 4x\) and \(g(x) = x - 7\), write \((g - f)(x)\) as a formula in \(x\).

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    \((x - 7) - 4x = -3x - 7\).

  9. 9.

    For \(f(x) = x + 1\) and \(g(x) = x^2 - 9\), find the domain of \(\left(\dfrac{f}{g}\right)(x)\), in interval notation.

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    Both functions accept every real number, but the quotient must exclude the zeros of \(g\): \(x^2 - 9 = 0\) at \(x = \pm 3\).

  10. 10.

    For \(f(x) = x^2\) and \(g(x) = x + 1\), write \(f(g(x))\) expanded, in terms of \(x\).

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    \(f(g(x)) = (x + 1)^2 = x^2 + 2x + 1\). The whole input is squared, so the bracket is essential.

  11. 11.

    For \(f(x) = x^2\) and \(g(x) = x + 1\), write \(g(f(x))\), in terms of \(x\).

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    \(g(f(x)) = g(x^2) = x^2 + 1\). Different from \(f(g(x))\) — composition is not commutative.

  12. 12.

    For \(f(x) = 2x + 5\) and \(g(x) = x - 3\), write \(f(g(x))\) simplified, in terms of \(x\).

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    \(f(g(x)) = 2(x - 3) + 5 = 2x - 6 + 5 = 2x - 1\).

  13. 13.

    For \(f(x) = \sqrt{x}\) and \(g(x) = x - 6\), find the domain of \(f(g(x))\), in interval notation.

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    \(f(g(x)) = \sqrt{x - 6}\), which needs \(x - 6 \ge 0\), so \(x \ge 6\). The outer function is what restricts here.

  14. 14.

    For \(f(x) = x^2 - 1\) and \(g(x) = x - 1\), simplify \(\left(\dfrac{f}{g}\right)(x)\) and give the simplified formula in terms of \(x\).

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    \(\dfrac{(x-1)(x+1)}{x-1} = x + 1\) — but \(x = 1\) is still excluded from the domain. Simplifying never restores a lost restriction.

  15. 15.

    A shop takes 20% off a price, then adds 5% tax on the reduced amount. What does a \(\pounds 50\) item cost, in pounds?

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    \(d(p) = 0.8p\) and \(t(u) = 1.05u\), so \(t(d(p)) = 1.05(0.8p) = 0.84p\). At \(p = 50\) that is \(\pounds 42\).
M
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Function Operations and Composition Practice Problems

Two lessons feed this set: Combining Functions and Function Composition. They are practised together because textbooks test them together, and because the fastest way to stop confusing with is to meet both in the same sitting.

How to use this set

There are two distinct operations here and it is worth naming which one you are doing before you start each problem.

Combining evaluates both functions at the same input and then does arithmetic on the two results.

Composing runs one function and feeds its output into the other as an input.

Evaluate compositions inside out, in two written steps. Trying to do both at once is where the bracket errors come from.

Common mistakes to watch for

Dropping the bracket when subtracting. is .

Running the outer function first. In , goes first.

Assuming . They differ in almost every case — problems 4 and 5 are the same two functions in opposite orders.

Forgetting the extra condition on a quotient. Zeros of leave the domain.

Letting a restriction vanish on simplification. It survives the algebra.

Where to go next

Function Notation is the evaluation skill these build on, Domain and Range goes deeper on the restrictions, and Inverse Functions uses composition as its verification step.

Frequently Asked Questions

What does (f + g)(x) mean?+

It means : evaluate both functions at the same input and add the results. The brackets on the left name a new function; they are not multiplication.

What is the difference between (fg)(x) and f(g(x))?+

multiplies the two outputs. is composition — the output of becomes the input of . They give different answers, and mixing them up is the most common error on this topic.

Which function runs first in f(g(x))?+

The inner one, , even though is written on the left. It is the same reading order as brackets in arithmetic: innermost first.

What is the domain of a combined function?+

The intersection of the two domains. For a quotient you must also exclude every input where .

What is the domain of a composition?+

The input must be in the domain of the inner function, **and** the value it produces must be in the domain of the outer one. Checking only the simplified formula can miss the inner restriction.

Does simplifying remove a domain restriction?+

No. simplifies to , but was never allowed and still is not — the original had a hole there.

More practice problems

Function Notation Practice Problems Inverse Functions Practice Problems Domain and Range Practice Problems