Composite functions
Algebra
- Higher
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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$f(x) = 2x + 1$ and $g(x) = x - 4$. Find $fg(3)$.
2 marks
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$f(x) = 2x + 1$ and $g(x) = x - 4$. Find $gf(x)$.
2 marks
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$f(x) = x^{2}$ and $g(x) = x + 3$. Find $fg(x)$ and $gf(x)$.
3 marks
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$f(x) = 3x - 2$. Find $f^{-1}(x)$.
3 marks
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$f(x) = 2x + 5$ and $g(x) = x^{2} - 1$. Find $fg(2)$, $gf(2)$ and $fg(x)$.
4 marks
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$f(x) = \dfrac{x - 1}{2}$. Find $ff(x)$ and $f^{-1}(x)$.
4 marks
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$f(x) = x + 4$ and $g(x) = x^{2}$. Find $fg(x)$, then solve $fg(x) = gf(x)$.
5 marks
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$f(x) = 2x - 1$ and $g(x) = \dfrac{1}{x + 3}$, $x \ne -3$. Find $fg(x)$ and $g^{-1}(x)$. Then solve $g(x) = \dfrac{1}{5}$.
6 marks
Mark scheme
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Answer. $-1$
- $g(3) = -1$
- $f(-1) = -1$
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Answer. $2x - 3$
- $f(x) = 2x + 1$
- $g(2x + 1) = 2x - 3$
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Answer. $(x + 3)^{2}$ and $x^{2} + 3$
- $fg(x) = (x + 3)^{2}$
- $gf(x) = x^{2} + 3$
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Answer. $f^{-1}(x) = \dfrac{x + 2}{3}$
- $y = 3x - 2$, so $x = \dfrac{y + 2}{3}$.
- $f^{-1}(x) = \dfrac{x + 2}{3}$
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Answer. $11$, $80$ and $2x^{2} + 3$
- $g(2) = 3$, so $fg(2) = 11$.
- $f(2) = 9$, so $gf(2) = 80$.
- $fg(x) = 2(x^{2} - 1) + 5 = 2x^{2} + 3$.
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Answer. $ff(x) = \dfrac{x - 3}{4}$ and $f^{-1}(x) = 2x + 1$
- $ff(x) = \dfrac{\frac{x - 1}{2} - 1}{2} = \dfrac{x - 3}{4}$.
- $f^{-1}(x) = 2x + 1$.
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Answer. $fg(x) = x^{2} + 4$, and $x = -1.5$
- $fg(x) = x^{2} + 4$ and $gf(x) = (x + 4)^{2}$.
- $x^{2} + 4 = x^{2} + 8x + 16$
- $8x = -12$, so $x = -1.5$.
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Answer. $fg(x) = -\dfrac{x + 1}{x + 3}$, $g^{-1}(x) = \dfrac{1}{x} - 3$, and $x = 2$
- $fg(x) = \dfrac{2}{x + 3} - 1 = -\dfrac{x + 1}{x + 3}$.
- From $y = \dfrac{1}{x + 3}$, $g^{-1}(x) = \dfrac{1}{x} - 3$.
- $\dfrac{1}{x + 3} = \dfrac{1}{5}$, so $x = 2$.