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Pure mathematics: advanced

Quick questions on Functions and the modulus function: domain, inverses, composites and modulus - OCR A-Level Maths A

7short Q&A pairs drawn directly from our worked dot-point answer. For full context and worked exam questions, read the parent dot-point page.

What are composite functions?
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The composite fg(x)fg(x) means "do gg first, then ff": fg(x)=f(g(x))fg(x) = f(g(x)). Order matters, so in general fg≠gffg \ne gf. The domain of fgfg is the set of inputs for which g(x)g(x) is itself a valid input to ff.
What are inverse functions?
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The inverse f−1f^{-1} undoes ff: if f(a)=bf(a) = b then f−1(b)=af^{-1}(b) = a. To find it, write y=f(x)y = f(x), swap the roles and solve for the new output. The graph of f−1f^{-1} is the reflection of the graph of ff in the line y=xy = x, and the domain and range swap.
What is the modulus function?
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The modulus ∣x∣|x| is the distance of xx from zero, so it is never negative: ∣x∣=x|x| = x for x≥0x \ge 0 and ∣x∣=−x|x| = -x for x<0x < 0. The graph y=∣f(x)∣y = |f(x)| takes the graph of ff and reflects any part below the xx-axis up above it.
What is restricting a domain for an inverse?
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A many-to-one function such as f(x)=x2f(x) = x^2 has no inverse over all reals, because two inputs give the same output. Restricting the domain to x≥0x \ge 0 makes it one-to-one, and then f−1(x)=xf^{-1}(x) = \sqrt{x}.
What are solving modulus problems?
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To solve ∣f(x)∣=g(x)|f(x)| = g(x) algebraically, square both sides (valid because both sides are then squares) or split into the two cases f(x)=g(x)f(x) = g(x) and f(x)=−g(x)f(x) = -g(x), checking each solution. A sketch confirms how many solutions there are and which region satisfies an inequality.
What is q1?
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Given f(x)=2x+5f(x) = 2x + 5, find f−1(x)f^{-1}(x). [2 marks]
What is q2?
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Solve ∣3x−2∣=7|3x - 2| = 7. [2 marks]

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