Hint
First factor f′. For the closed interval, evaluate both endpoints as well as the stationary inputs.
Worked solution
(a)
\(f^{\prime}=3x^2-12x+9=3(x-1)(x-3)\). The stationary inputs are 1 and 3. Their ordinates are \(\displaystyle f(1)=5\) and \(\displaystyle f(3)=1\), so the points are (1,5) and (3,1).
Answer: \((1,5)\) and \((3,1)\).
(b)
The signs of 3(x−1)(x−3) are +,−,+ on the three intervals. Thus f increases, then decreases, then increases. The positive-to-negative change at 1 gives a local maximum; the negative-to-positive change at 3 gives a local minimum.
Answer: Increasing for x<1 and x>3; decreasing for 1<x<3. Local maximum (1,5), local minimum (3,1).
(c)
The candidate values are \(\displaystyle f(0)=1, f(1)=5, f(3)=1\) and \(\displaystyle f(4)=5\). Continuity and the derivative sign pattern account for the whole closed interval. The maximum is 5 at both 1 and 4; the minimum is 1 at both 0 and 3.
Answer: Maximum 5 at \(\displaystyle x=1\) and 4; minimum 1 at \(\displaystyle x=0\) and 3.
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) · 3 marks
- 1 mark: Obtain and factor the correct first derivative.
- 1 mark: Find both stationary inputs 1 and 3.
- 1 mark: Give the corresponding points (1,5) and (3,1).
Alternative methods: Solving the derivative quadratic by the formula is accepted.
Common error: Giving derivative values as ordinates.
(b) · 3 marks
- 1 mark: Give both intervals with positive derivative.
- 1 mark: Give the negative-derivative interval.
- 1 mark: Classify the two points correctly with derivative-sign or valid second-derivative reasoning.
Alternative methods: Use f″\(\displaystyle (1)=-6\) and f″\(\displaystyle (3)=6\) for classification, together with correct monotonic intervals.
Common error: Reversing the middle sign or calling the stationary points absolute extrema on the whole real line.
(c) · 3 marks
- 1 mark: Evaluate both included endpoints and compare with the stationary values.
- 1 mark: Give maximum 5 with both \(\displaystyle x=1\) and \(\displaystyle x=4\).
- 1 mark: Give minimum 1 with both \(\displaystyle x=0\) and \(\displaystyle x=3\).
Alternative methods: A complete sign-chart and endpoint-value argument is accepted.
Common error: Omitting the endpoint that ties an interior extremum.
Total: 9 marks.