These eleven questions cover evaluating functions, composites, inverses, restricted domains and real contexts. They are original, ordered from easier to harder, and written for functions and mappings.
Try each one on paper first. Show every step, because in an exam the method earns credit as well as the final value. Open the answer only after you have an attempt.
Questions and worked answers
1. If f(x) = 4x − 7, find f(5) and f(−2).
Show answer
f(5) = 4(5) − 7 = 20 − 7 = 13. f(−2) = 4(−2) − 7 = −8 − 7 = −15.
2. If f(x) = x² + 3x, find f(−4).
Show answer
f(−4) = (−4)² + 3(−4) = 16 − 12 = 4.
3. If f(x) = 10 − x², find f(−3).
Show answer
f(−3) = 10 − (−3)² = 10 − 9 = 1.
4. If f(x) = 2x² − x + 1, find f(−2).
Show answer
f(−2) = 2(−2)² − (−2) + 1 = 2(4) + 2 + 1 = 8 + 2 + 1 = 11.
5. Let f(x) = x + 2 and g(x) = 3x. Find fg(4) and gf(4).
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fg(4): g(4) = 12, then f(12) = 14. So fg(4) = 14. gf(4): f(4) = 6, then g(6) = 18. So gf(4) = 18. The two orders give different answers.
6. Let f(x) = x² − 1 and g(x) = 2x + 3. Find fg(x) and gf(x), simplified.
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fg(x) = (2x + 3)² − 1 = 4x² + 12x + 9 − 1 = 4x² + 12x + 8. gf(x) = 2(x² − 1) + 3 = 2x² − 2 + 3 = 2x² + 1. Check with x = 1: fg(1) = f(5) = 24 and 4 + 12 + 8 = 24. gf(1) = g(0) = 3 and 2 + 1 = 3.
7. Find the inverse of f(x) = (x − 4)/3, then find f⁻¹(5).
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y = (x − 4)/3, so 3y = x − 4 and x = 3y + 4. Hence f⁻¹(x) = 3x + 4. f⁻¹(5) = 3(5) + 4 = 19. Check: f(19) = 15/3 = 5.
8. If f(x) = 7 − 3x, find f⁻¹(−5).
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y = 7 − 3x, so 3x = 7 − y and x = (7 − y)/3. So f⁻¹(x) = (7 − x)/3. f⁻¹(−5) = (7 + 5)/3 = 12/3 = 4. Check: f(4) = 7 − 12 = −5.
9. State the values of x for which f(x) = √(5 − 2x)/(x + 1) is defined.
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Square root: 5 − 2x ≥ 0, so 2x ≤ 5 and x ≤ 2.5. Denominator: x + 1 ≠ 0, so x ≠ −1. Both conditions together: x ≤ 2.5, x ≠ −1.
10. A mobile data plan charges RM20 plus RM0.80 per GB, so c = 20 + 0.8g, for 0 ≤ g ≤ 10 GB. (a) Find the cost for 6.5 GB. (b) A bill is RM26.80. How many GB were used? (c) Could a bill of RM30 be correct?
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(a) c = 20 + 0.8 × 6.5 = 20 + 5.2 = RM25.20. (b) 20 + 0.8g = 26.8, so 0.8g = 6.8 and g = 6.8 ÷ 0.8 = 8.5 GB. This is within 0 to 10. Check: 0.8 × 8.5 = 6.8. (c) 20 + 0.8g = 30 gives 0.8g = 10 and g = 12.5. That is above 10, so no, RM30 is not possible under this plan.
11. Let f(x) = 2x + 1 and g(x) = x². (a) Solve fg(x) = 51. (b) Solve gf(x) = 49.
Show answer
(a) fg(x) = 2x² + 1. So 2x² + 1 = 51, 2x² = 50, x² = 25, giving x = 5 or x = −5. Check: 2(25) + 1 = 51. (b) gf(x) = (2x + 1)². So (2x + 1)² = 49, giving 2x + 1 = 7 or 2x + 1 = −7. Then x = 3 or x = −4. Check x = −4: (−8 + 1)² = 49.
If you got these wrong
Match the kind of error to the lesson that teaches the fix.
| What went wrong | Questions | Go back to |
|---|---|---|
| Lost a sign or a bracket when substituting a negative | 1 to 4 | Evaluate a function with negative inputs |
| Applied the functions in the wrong order, or expanded a bracket wrongly | 5, 6, 11 | Trace an input through a composite function |
| Undid the operations in the wrong order, or took a reciprocal | 7, 8 | Reverse a one-to-one mapping |
| Forgot one of the conditions, or reversed an inequality | 9 | Restrict inputs to keep an expression defined |
| Stopped one step early, or ignored the domain and units | 10 | Interpret a function machine in context |
Use the function composition and inverse explorer to test any rule with your own numbers, and the mistake log and retest queue to keep track of which error types keep returning. The non-calculator working trainer is handy for the arithmetic steps.
If a whole row of this table looks familiar, our teachers can go through it with you in online one-to-one Mathematics tuition.