A-Level Functions and their graphs Practice Questions

Free A-Level Functions and their graphs practice questions with full step-by-step worked solutions. Covers functions, composite, inverse, modulus. Practise exam-style problems and check your method.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The functions are defined by f(x)=2x+1f(x)=2 x + 1 and g(x)=x−3g(x)=x - 3. Find fg(5)fg(5), where fg(x)=f(g(x))fg(x)=f(g(x)).
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Worked solution

  1. State the two functions

    f(x)=2x+1,g(x)=x−3f(x)=2 x + 1,\quad g(x)=x - 3

    We begin from the given definitions.

  2. Work out the inner function g(5)

    g(5)=2g(5)=2

    fg means f(g(x)), so evaluate the inner function first.

  3. State the value of fg(5)

    fg(5)=5fg(5)=5

    This is the required value.

Answer
55
Question 2
2 markseasy
State the range of f(x)=∣x−2∣f(x)=\left|{x - 2}\right|.
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Worked solution

  1. State the function

    f(x)=∣x−2∣f(x)=\left|{x - 2}\right|

    We begin from the given definition.

  2. Determine the possible output values

    ∣x−2∣≥0\left|x-2\right|\ge 0

    The range is the set of values the function can take.

  3. State the answer

    f(x)≥0f(x) \ge 0

    This is the range of the function.

Answer
f(x)≥0f(x) \ge 0
Question 3
3 marksintermediate
State the range of f(x)=3−x2f(x)=3 - x^{2}.
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Worked solution

  1. State the function

    f(x)=3−x2f(x)=3 - x^{2}

    We begin from the given definition.

  2. Determine the possible output values

    −x2≤0 so 3−x2≤3-x^2\le 0 \text{ so } 3-x^2\le 3

    The range is the set of values the function can take.

  3. Locate the turning value

    maximum at x=0\text{maximum at } x=0

    The extreme value bounds the range.

  4. Consider the behaviour for large |x|

    examine the outputs as x→±∞\text{examine the outputs as } x \to \pm\infty

    The end behaviour helps fix the extent of the range.

  5. Combine the bound with the reachable values

    −x2≤0 so 3−x2≤3-x^2\le 0 \text{ so } 3-x^2\le 3

    The extreme value together with the trend fixes the range.

  6. State the answer

    f(x)≤3f(x) \le 3

    This is the range of the function.

Answer
f(x)≤3f(x) \le 3
Question 4
5 markshard
Given f(x)=1x−1f(x)=\frac{1}{x - 1} and g(x)=2x+3g(x)=2 x + 3, find fg(x)=f(g(x))fg(x)=f(g(x)).
Show worked solution

Worked solution

  1. State the two functions

    f(x)=1x−1,g(x)=2x+3f(x)=\frac{1}{x - 1},\quad g(x)=2 x + 3

    We begin from the given definitions.

  2. Substitute the inner function into the outer

    fg(x)=f(g)fg(x)=f(g)

    Replace the variable of the outer function by the inner function.

  3. Write out the substitution explicitly

    fg(x)=12x+2fg(x)=\frac{1}{2 x + 2}

    Put the inner expression wherever x appears in the outer function.

  4. Note the domain of the composite

    require x in the domain of g\text{require } x \text{ in the domain of } g

    The composite is only defined where the inner function is.

  5. Substitute x=-3 as a numerical check

    x=−3: −14x=-3:\ - \frac{1}{4}

    Evaluating at a point provides a check on the result.

  6. Substitute x=-2 as a numerical check

    x=−2: −12x=-2:\ - \frac{1}{2}

    Evaluating at a point provides a check on the result.

  7. Substitute x=0 as a numerical check

    x=0: 12x=0:\ \frac{1}{2}

    Evaluating at a point provides a check on the result.

  8. Substitute x=1 as a numerical check

    x=1: 14x=1:\ \frac{1}{4}

    Evaluating at a point provides a check on the result.

  9. Substitute x=2 as a numerical check

    x=2: 16x=2:\ \frac{1}{6}

    Evaluating at a point provides a check on the result.

  10. State the composite function

    fg(x)=12(x+1)fg(x)=\frac{1}{2 \left(x + 1\right)}

    This is the required composite function.

Answer
12(x+1)\frac{1}{2 \left(x + 1\right)}
Question 5
8 markschallenging
State the range of f(x)=1x2+1f(x)=\frac{1}{x^{2} + 1}.
Show worked solution

Worked solution

  1. State the function

    f(x)=1x2+1f(x)=\frac{1}{x^{2} + 1}

    We begin from the given definition.

  2. Determine the possible output values

    x2+1≥1 so 0<1x2+1≤1x^2+1\ge 1 \text{ so } 0<\tfrac{1}{x^2+1}\le 1

    The range is the set of values the function can take.

  3. Locate the turning value

    maximum at x=0\text{maximum at } x=0

    The extreme value bounds the range.

  4. Consider the behaviour for large |x|

    examine the outputs as x→±∞\text{examine the outputs as } x \to \pm\infty

    The end behaviour helps fix the extent of the range.

  5. Combine the bound with the reachable values

    x2+1≥1 so 0<1x2+1≤1x^2+1\ge 1 \text{ so } 0<\tfrac{1}{x^2+1}\le 1

    The extreme value together with the trend fixes the range.

  6. Substitute x=-3 as a numerical check

    x=−3: 110x=-3:\ \frac{1}{10}

    Evaluating at a point provides a check on the result.

  7. Substitute x=-2 as a numerical check

    x=−2: 15x=-2:\ \frac{1}{5}

    Evaluating at a point provides a check on the result.

  8. Substitute x=-1 as a numerical check

    x=−1: 12x=-1:\ \frac{1}{2}

    Evaluating at a point provides a check on the result.

  9. Substitute x=0 as a numerical check

    x=0: 1x=0:\ 1

    Evaluating at a point provides a check on the result.

  10. Substitute x=1 as a numerical check

    x=1: 12x=1:\ \frac{1}{2}

    Evaluating at a point provides a check on the result.

  11. Substitute x=2 as a numerical check

    x=2: 15x=2:\ \frac{1}{5}

    Evaluating at a point provides a check on the result.

  12. Substitute x=3 as a numerical check

    x=3: 110x=3:\ \frac{1}{10}

    Evaluating at a point provides a check on the result.

  13. Substitute x=4 as a numerical check

    x=4: 117x=4:\ \frac{1}{17}

    Evaluating at a point provides a check on the result.

  14. Substitute x=5 as a numerical check

    x=5: 126x=5:\ \frac{1}{26}

    Evaluating at a point provides a check on the result.

  15. State the answer

    0<f(x)≤10 < f(x) \le 1

    This is the range of the function.

Answer
0<f(x)≤10 < f(x) \le 1

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