This set has twelve original questions, ordered from easier to harder, covering all five lessons in functions and restrictions. Questions 1 to 3 are warm-ups on domains, 4 and 5 are ranges, 6 and 7 are composites, 8 and 9 are inverses, and 10 to 12 mix skills.
Attempt each question on paper and write every restriction as its own line, as you would in an exam. Only then open the answer. Mark the ones you got wrong and use the routing list at the end.
The function composition and inverse explorer is useful for checking an answer after you have tried by hand.
Questions
1. State the largest possible domain of f(x) = 4/(x − 5).
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The denominator is zero when x − 5 = 0, so x = 5.
Domain: all real x with x ≠ 5.
2. State the largest possible domain of g(x) = (x + 2)/(x² − 9).
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x² − 9 = (x − 3)(x + 3) = 0 gives x = 3 or x = −3. The numerator is ignored.
Domain: all real x with x ≠ 3 and x ≠ −3.
3. State the largest possible domain of h(x) = 2x/(x² + x − 12).
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x² + x − 12 = (x + 4)(x − 3) = 0 gives x = −4 or x = 3. Check: 16 − 4 − 12 = 0 and 9 + 3 − 12 = 0.
Domain: all real x with x ≠ −4 and x ≠ 3.
4. The function f is defined by f(x) = 3x − 2 for −1 ≤ x ≤ 4. Find the range of f.
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f is a straight line that increases, so the endpoints give the extremes. f(−1) = −3 − 2 = −5 and f(4) = 12 − 2 = 10.
Range: −5 ≤ f(x) ≤ 10.
5. The function f is defined by f(x) = x² − 6x + 2 for 0 ≤ x ≤ 5. Find the range of f.
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Complete the square: f(x) = (x − 3)² − 7. The minimum is −7 at x = 3, which lies inside 0 ≤ x ≤ 5.
Endpoints: f(0) = 2 and f(5) = 25 − 30 + 2 = −3. Check with the square: (0 − 3)² − 7 = 2 and (5 − 3)² − 7 = −3. The largest is 2.
Range: −7 ≤ f(x) ≤ 2.
6. f(x) = 2x − 3 and g(x) = x² + 1. Find fg(x) and gf(x).
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fg(x) = f(x² + 1) = 2(x² + 1) − 3 = 2x² − 1.
gf(x) = g(2x − 3) = (2x − 3)² + 1 = 4x² − 12x + 9 + 1 = 4x² − 12x + 10.
Check with x = 1: g(1) = 2 and f(2) = 1, and 2 − 1 = 1. Also f(1) = −1 and g(−1) = 2, and 4 − 12 + 10 = 2.
fg(x) = 2x² − 1 and gf(x) = 4x² − 12x + 10.
7. Using the functions in question 6, solve fg(x) = 17.
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2x² − 1 = 17, so 2x² = 18, so x² = 9.
x = 3 or x = −3. Check x = 3: g(3) = 10 and f(10) = 17. Check x = −3: g(−3) = 10 as well.
8. The function f is defined by f(x) = (3x − 1)/(x + 2) for x ≠ −2. Find f⁻¹(x) and state its domain.
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y(x + 2) = 3x − 1, so yx + 2y = 3x − 1, so yx − 3x = −1 − 2y, so x(y − 3) = −(1 + 2y), so x = (2y + 1)/(3 − y).
f⁻¹(x) = (2x + 1)/(3 − x).
Range of f: f(x) = 3 − 7/(x + 2), which never equals 3. So the domain of f⁻¹ is x ≠ 3.
Check: f(1) = 2/3, and f⁻¹(2/3) = (4/3 + 1)/(3 − 2/3) = (7/3)/(7/3) = 1.
9. The function f is defined by f(x) = x² − 2 for x ≥ 0. Find f⁻¹(x) and state its domain.
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y = x² − 2, so x² = y + 2, and since x ≥ 0, x = √(y + 2).
f⁻¹(x) = √(x + 2). The range of f is f(x) ≥ −2, so the domain of f⁻¹ is x ≥ −2.
Check: f(3) = 7, and f⁻¹(7) = √9 = 3.
10. The function f is defined by f(x) = x² − 8x + 3 for real x. Show that f has no inverse. Then find the largest domain of the form x ≥ k for which f⁻¹ exists, and find f⁻¹(x).
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f(0) = 3 and f(8) = 64 − 64 + 3 = 3. Two different inputs give the same output, so f is many-to-one and has no inverse on all real x.
Complete the square: f(x) = (x − 4)² − 13. The turning point is at x = 4, so k = 4. On x ≥ 4 the range is f(x) ≥ −13.
y = (x − 4)² − 13 gives x − 4 = √(y + 13), so x = 4 + √(y + 13).
f⁻¹(x) = 4 + √(x + 13) for x ≥ −13. Check: f(6) = 36 − 48 + 3 = −9, and f⁻¹(−9) = 4 + 2 = 6.
11. f(x) = 1/(x − 2) for x ≠ 2, and g(x) = x + 5 for all real x. Find fg(x) and gf(x), and state the values of x each must exclude.
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fg(x) = f(x + 5) = 1/(x + 5 − 2) = 1/(x + 3). The denominator is zero at x = −3, so x ≠ −3.
gf(x) = g(1/(x − 2)) = 1/(x − 2) + 5. The inner function f excludes x = 2, so x ≠ 2.
12. The function f is defined by f(x) = x² + 4x for x ≥ −2. Find the range of f, then find f⁻¹(x) and use it to evaluate f⁻¹(5).
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Complete the square: f(x) = (x + 2)² − 4. The minimum is −4 at x = −2, and f increases for x ≥ −2. Range: f(x) ≥ −4.
y = (x + 2)² − 4 gives x + 2 = √(y + 4), so x = −2 + √(y + 4).
f⁻¹(x) = −2 + √(x + 4) for x ≥ −4.
f⁻¹(5) = −2 + √9 = −2 + 3 = 1. Check: f(1) = 1 + 4 = 5.
If you got these wrong
- Questions 1 to 3, wrong or missing excluded values: return to finding valid inputs for a rational expression. Set the original denominator to zero and do not cancel first.
- Questions 4 and 5, range out by a turning point: return to determining a range from a restricted domain. Always ask whether a turning point lies inside the interval.
- Questions 6, 7 and 11, wrong order or missing exclusion: return to forming a composite function in the correct order. Read fg(x) as f(g(x)) and check with one number.
- Questions 8, 9 and 12, inverse or its domain: return to finding an inverse and checking its domain. The domain of f⁻¹ is the range of f.
- Question 10, no inverse or wrong restriction: return to explaining a many-to-one mapping. Start from the turning point.
Record each error type in the mistake log and retest queue, then retry a fresh question on the same skill a few days later. The non-calculator working trainer helps with checking expanded brackets and exact values.
If one error type keeps coming back, it is usually a habit and not a gap in understanding. A teacher can find it quickly in online one-to-one Additional Mathematics tuition.