Hint
Compare the two endpoint outputs. The inverse domain is the original range.
Worked solution
(a)
The function decreases: \(f(-1)=8\) and \(f(4)=-7\). Every intermediate value is attained, so the range is \(-7\le y\le8\).
Answer: \([-7,8]\).
(b)
From \(y=5-3x\), \(x=(5-y)/3\). Exchange the names to obtain the inverse; its domain is the range from part (a).
Answer: \(f^{-1}(x)=(5-x)/3\), \(-7\le x\le8\).
(c)
On the common domain \([-1,4]\), solve \((5-x)/3=5-3x\). This gives \(8x=10\), hence \(x=5/4\), which is permitted. Both outputs are \(5/4\).
Answer: \(x=5/4\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Calculate both endpoint outputs −7 and 8.
- 1 mark: Give the inclusive interval [−7,8] in the correct order.
Alternative methods: A monotonicity argument with endpoint substitution is sufficient.
Common error: Keeping the endpoint order of the inputs after a decreasing mapping.
(b) · 3 marks
- 1 mark: Correctly rearrange y=5−3x for x.
- 1 mark: State (5−x)/3 as the inverse.
- 1 mark: State the inverse domain [−7,8].
Alternative methods: A direct reversal of the arithmetic operations is accepted.
Common error: Using [−1,4] as the inverse domain.
(c) · 3 marks
- 1 mark: Equate the two correct expressions.
- 1 mark: Solve to x=5/4.
- 1 mark: Check the value lies in both domains.
Alternative methods: Solving f(x)=x is accepted here if the equivalence for this particular linear function is justified.
Common error: Ignoring the domain check or assuming a general inverse-intersection rule.
Total: 8 marks.