[discia] GCSE Maths › Algebra › Inverse and Composite Functions › Practise
  1. Either paper 4 marks

    \(f(x) = 2x - 3\) and \(g(x) = 4x - 7\)

    (a)

    Find the value of \(fg(2)\).

    (b)

    Find the value of \(gf(2)\).

    Show answer
    (a) \(-1\)
    (b) \(-3\)

    \(fg(2)\) means do \(g\) first, then put its answer into \(f\).

    \(g(2) = 4(2) - 7 = 1\)

    \(f(1) = 2(1) - 3 = -1\)

    This time \(f\) goes first.

    \(f(2) = 2(2) - 3 = 1\)

    \(g(1) = 4(1) - 7 = -3\)

    Notice \(fg(2) = -1\) but \(gf(2) = -3\). The order the two functions are applied in changes the answer, which is why the notation is worth reading carefully.

    • (a) M1g(2) = \(1\)
    • (a) A1fg(2) = \(-1\)
    • (b) M1f(2) = \(1\)
    • (b) A1gf(2) = \(-3\)
  2. Either paper 3 marks

    \(f(x) = ax - 5\)

    \(f(3) = 13\)

    Work out the value of \(a\).

    Show answer
    \(6\)

    \(f(3)\) means put \(3\) in place of \(x\).

    \(a \times 3 - 5 = 13\)

    \(3a = 13 + 5\)

    \(3a = 18\), so \(a = 18 \div 3 = 6\)

    • M1substituting \(x = 3\), \(3a - 5 = 13\)
    • M1isolating the term, \(3a = 18\)
    • A1\(a = 6\)
  3. Either paper 3 marks

    \(f(x) = 7x + 5\) and \(g(x) = 6x - 5\)

    Find \(fg(x)\).

    Give your answer in the form \(px + q\), where \(p\) and \(q\) are integers.

    Show answer
    \(42x - 30\)

    \(fg(x)\) means substitute the whole of \(g(x)\) in place of \(x\) in \(f\).

    \(fg(x) = f(6x - 5) = 7(6x - 5) + 5\)

    Expand the bracket:

    \(fg(x) = 42x - 35 + 5\)

    Collect the two numbers:

    \(fg(x) = 42x - 30\)

    So \(p = 42\) and \(q = -30\).

    • M1substituting \(g(x)\) into \(f\), \(7(6x - 5) + 5\)
    • M1expanding \(42x - 35 + 5\)
    • A1\(42x - 30\)
  4. Either paper 3 marks

    \(f(x) = 6x - 6\)

    Find an expression for \(f^{-1}(x)\).

    Show answer
    \(f^{-1}(x) = \frac{x + 6}{ 6 }\)

    Write the function as an equation. Use \(z\) for the output, so the letter \(x\) is free to become the subject.

    \(z = 6x - 6\)

    Undo the function step by step. Add \(6\) to both sides — the opposite of the \(- 6\):

    \(z + 6 = 6x\)

    Divide both sides by \(6\):

    \(\frac{z + 6}{ 6 } = x\)

    Now swap \(z\) back to \(x\), because the inverse is written as a function of its own input:

    \(f^{-1}(x) = \frac{x + 6}{ 6 }\)

    • M1rearranging \(z + 6 = 6x\)
    • M1dividing by 6, \(x = \frac{z + 6}{ 6 }\)
    • A1\(f^{-1}(x) = \frac{x + 6}{ 6 }\)
  5. Either paper 3 marks

    \(f(x) = 7x - 6\) and \(g(x) = x + 4\)

    \(fg(k) = 50\)

    Find the value of \(k\).

    Show answer
    \(4\)

    \(fg(k)\) means work out \(g(k)\) first, then put that into \(f\).

    \(g(k) = k + 4\)

    \(f(k + 4) = 7(k + 4) - 6 = 50\)

    \(7(k + 4) = 56\)

    \(k + 4 = 8\), so \(k = 4\)

    • M1substitutes \(g(k)\) into \(f\): \(7(k + 4) - 6 = 50\)
    • M1isolates the bracket: \(k + 4 = 8\)
    • A1k = \(4\)
  6. Either paper 4 marks

    \(f(x) = 5x + 6\)

    (a)

    Work out \(f^{-1}(x)\).

    (b)

    Solve the equation \(f^{-1}(x) = -3\).

    Show answer
    (a) \(f^{-1}(x) = \frac{x - 6}{ 5 }\)
    (b) \(-9\)

    Put \(z = 5x + 6\) and make \(x\) the subject.

    \(z - 6 = 5x\), so \(x = \frac{z - 6}{ 5 }\).

    Written as a function of its own input, \(f^{-1}(x) = \frac{x - 6}{ 5 }\).

    Using the inverse from part (a):

    \(\frac{x - 6}{ 5 } = -3\)

    \(x - 6 = 5(-3) = -15\)

    \(x = -9\)

    There is a quicker route worth seeing. \(f^{-1}(x) = -3\) says exactly the same thing as \(x = f(-3)\), so \(x = 5(-3) + 6 = -9\) in one line.

    • (a) M1rearranges to \(x = \frac{z - 6}{ 5 }\)
    • (a) A1f^{-1}(x) = \(\frac{x - 6}{ 5 }\)
    • (b) M1x - 6 = \(-15\), or x = f(\(-3\))
    • (b) A1x = \(-9\)
  7. Either paper 5 marks

    \(f(x) = x^2 - 1\) and \(g(x) = x - 3\)

    (a)

    Work out \(gf(x)\).

    (b)

    Work out \(fg(x)\).

    Give your answer in the form \(x^2 + px + q\).

    Show answer
    (a) \(x^2 - 4\)
    (b) \(x^2 - 6x + 8\)

    \(gf(x)\) means \(f\) first, then \(g\).

    \(f(x) = x^2 - 1\), and \(g\) adds \(-3\) to whatever it is given:

    \(gf(x) = (x^2 - 1) - 3 = x^2 - 4\)

    This time \(g\) goes first, so the whole of \(g(x)\) is squared.

    \(fg(x) = (x - 3)^2 - 1\)

    Expand the bracket:

    \((x - 3)^2 = x^2 - 6x + 9\)

    So \(fg(x) = x^2 - 6x + 9 - 1 = x^2 - 6x + 8\).

    Compare it with part (a): squaring after adding is not the same as adding after squaring, and here the difference is a whole \(x\) term.

    • (a) M1substitutes \(f(x)\) into \(g\): \((x^2 - 1) - 3\)
    • (a) A1\(gf(x) = x^2 - 4\)
    • (b) M1substitutes \(g(x)\) into \(f\): \((x - 3)^2 - 1\)
    • (b) M1expands the bracket: \(x^2 - 6x + 9\)
    • (b) A1\(fg(x) = x^2 - 6x + 8\)
  8. Either paper 4 marks

    The graph of \(y = h(x)\) is drawn on the grid.

    xy-7-6-5-4-3-21234567-7-6-5-4-3-212345670
    (a)

    On the grid, draw the graph of the inverse function \(y = h^{-1}(x)\) for \(1 \le x \le 3\). Plot its two end points, then join them.

    (b)

    \(f(x) = \sin x\) and \(g(x) = x + 90\)

    On the grid, draw the graph of the composite function \(y = fg(x)\) for \(0 \le x \le 360\). Plot its points at \(x = 0\), \(90\), \(180\), \(270\) and \(360\), then draw the curve.

    xy90180270360-1-0.50.510

    You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.

    Show answer
    (a) The line segment from \((1, 0)\) to \((3, 1)\), with gradient \(\frac{1}{2}\)
    (b) The sine curve \(y = \sin(x + 90)\) through \((0, 1)\), \((90, 0)\), \((180, -1)\), \((270, 0)\) and \((360, 1)\)
    (a)xy-7-6-5-4-3-21234567-7-6-5-4-3-212345670
    (b)xy90180270360-1-0.50.510

    The graph of an inverse function is the reflection of the graph of the function in the line \(y = x\). Reflecting in \(y = x\) swaps the \(x\)- and \(y\)-coordinates of every point.

    \((0, 1) \to (1, 0)\)

    \((1, 3) \to (3, 1)\)

    Plot these two points and join them with a straight line.

    Check the gradient: \(h\) has gradient \(2\), going up 2 for every 1 across. So \(h^{-1}\) goes up 1 for every 2 across, which is a gradient of \(\frac{1}{2}\), and the line you drew should do this.

    In \(fg(x)\) the function \(g\) is applied first, so

    \(fg(x) = f(x + 90) = \sin(x + 90)\)

    This is the graph of \(y = \sin x\) moved 90 degrees to the left. Find its values at the marked \(x\)-values:

    \(x = 0\): \(\sin 90 = 1\)

    \(x = 90\): \(\sin 180 = 0\)

    \(x = 180\): \(\sin 270 = -1\)

    \(x = 270\): \(\sin 360 = 0\)

    \(x = 360\): \(\sin 450 = 1\)

    Plot these five points and join them with a smooth curve.

    • (a) B1one end point plotted, \((1, 0)\) or \((3, 1)\)
    • (a) B1both end points \((1, 0)\) and \((3, 1)\) plotted and joined
    • (b) M1fg(x) = sin(x + 90), the sine curve moved 90 degrees to the left, shown by at least three of the five points
    • (b) A1smooth curve through \((0, 1)\), \((90, 0)\), \((180, -1)\), \((270, 0)\), \((360, 1)\)
  9. Either paper 4 marks

    \(f(x) = 5x - 4\)

    Show that \(4f(x) - 100f^{-1}(x)\) simplifies to an integer.

    Show answer
    The \(x\) terms cancel and \(4f(x) - 100f^{-1}(x) = -96\), an integer.

    Find the inverse. Write \(z = 5x - 4\) and make \(x\) the subject:

    \(x = \frac{z + 4}{ 5 }\), so \(f^{-1}(x) = \frac{x + 4}{ 5 }\)

    Multiply \(f(x)\) by \(4\) and expand:

    \(4f(x) = 4(5x - 4) = 20x - 16\)

    Multiply \(f^{-1}(x)\) by \(100\). Since \(100 \div 5 = 20\):

    \(100f^{-1}(x) = 20(x + 4) = 20x + 80\)

    Subtract, taking care with the signs in the second bracket:

    \(4f(x) - 100f^{-1}(x) = (20x - 16) - (20x + 80)\)

    \(= 20x - 16 - 20x - 80\)

    The \(x\) terms cancel, leaving

    \(-16 - 80 = -96\)

    This is an integer, so the result is shown.

    • M1inverse \(f^{-1}(x) = \frac{x + 4}{ 5 }\)
    • M1\(4f(x) = 20x - 16\) expanded
    • M1\(100f^{-1}(x) = 20x + 80\), with \(100\) divided by \(5\) and both terms of the bracket multiplied
    • A1subtracts, the \(x\) terms cancelling, to reach \(-96\)
  10. Either paper 4 marks

    \(f(x) = x^2\) and \(g(x) = -8x - 12\)

    Solve \(f(x) = g(x)\).

    Show answer
    \(x = -6\) or \(x = -2\)

    Write down what the question says, then get everything onto one side.

    \(x^2 = -8x - 12\)

    \(x^2 + 8x + 12 = 0\)

    Factorise. Two numbers that multiply to \(12\) and add to \(8\) are \(6\) and \(2\):

    \((x + 6)(x + 2) = 0\)

    A product is zero only when one of its factors is, so \(x = -6\) or \(x = -2\).

    • M1rearranging to \(x^2 + 8x + 12 = 0\)
    • M1factorising \((x + 6)(x + 2) = 0\)
    • A1one solution \(x = -6\)
    • A1other solution \(x = -2\)