Edexcel International A Level Maths · P2 · Pure Mathematics 2
Stationary points and optimisation
28 free, high-quality past-paper-style questions on Stationary points and optimisation for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 243 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P2-7-B-01P2 · ASEasy6 marks
A cubic curve C is defined for every real value of x by y = x3 − 12x + 5. It has two turning points.
(a)
Find dy/dx.
(1)
(b)
Find the coordinates of the stationary points of C.
(3)
(c)
Use the second derivative to determine the nature of each stationary point.
(2)
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Answer
Mark
Notes
Part (a)
3x2 − 12
B1
Part (b)
sets 3x2 − 12 = 0
M1
x = 2 and x = −2
A1
(2, −11) and (−2, 21)
A1
Part (c)
d2y/dx2 = 6x, substituting both x-values
M1
at x = 2: 12 > 0 so (2, −11) is a minimum; at x = −2: −12 < 0 so (−2, 21) is a maximum
A1
Model sketch
Figure: model sketch of C with its stationary points
Common error: Finding only x = 2 from x2 = 4. A square has two square roots; the negative one gives the second stationary point.
The function f is defined by f(x) = 2x3 + 3x2 − 36x + 7, x ∈ ℝ.
(a)
Find f′(x), giving your answer in fully factorised form.
(2)
(b)
Find the set of values of x for which f is increasing.
(3)
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Answer
Mark
Notes
Part (a)
6x2 + 6x − 36
B1
6(x + 3)(x − 2)
A1
Part (b)
critical values x = −3 and x = 2
M1
chooses the outside region, since f′(x) > 0 outside the roots of an upward parabola
M1
x < −3 or x > 2
A1
Exam tip: “Increasing” is a condition on the gradient: solve f′(x) > 0. A quick sketch of the parabola f′ shows at once which side of the critical values you need.
A rectangular vegetable plot is enclosed on all four sides by 46 m of edging. The length of the plot is x metres and its area is A m2.
Figure: the vegetable plot (not to scale)
(a)
Show that A = 23x − x2.
(2)
(b)
Use calculus to find the maximum area of the plot, and justify that it is a maximum.
(3)
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Answer
Mark
Notes
Part (a)
width = (46 − 2x)/2 = 23 − x
M1
A = x(23 − x) = 23x − x2 as required
A1
Part (b)
dA/dx = 23 − 2x = 0
M1
x = 11.5, A = 132.25 m2
A1
d2A/dx2 = −2 < 0, so maximum
B1
Exam tip: When the question says “use calculus”, completing the square is not enough: show the derivative, set it equal to zero, and justify with the second derivative.
Figure 1 shows part of the curve C with equation y = 2x3 − 9x2 + 12x − 3, for x ≥ 0. The points P and Q are the two stationary points of C.
(a)
Find dy/dx and hence find the coordinates of P and of Q.
(4)
(b)
Use the second derivative to verify that P is a maximum point and Q is a minimum point.
(2)
(c)
Write down the set of values of x for which y is decreasing.
(1)
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Answer
Mark
Notes
Part (a)
dy/dx = 6x2 − 18x + 12
B1
6(x − 1)(x − 2) = 0
M1
x = 1 and x = 2
A1
P(1, 2) and Q(2, 1)
A1
Part (b)
d2y/dx2 = 12x − 18, substituting both x-values
M1
at x = 1: −6 < 0 so P is a maximum; at x = 2: 6 > 0 so Q is a minimum
A1
Part (c)
1 < x < 2 ft their P and Q
B1
Exam tip: The figure tells you which point should be the maximum, but the second derivative is what proves it. Write the value of d2y/dx2 at each point and state its sign before giving the conclusion.
The curve C has equation y = f(x). Figure 1 shows the graph of the gradient function y = f′(x), which is a quadratic. The graph of y = f′(x) crosses the x-axis at (−1, 0) and (3, 0) and crosses the y-axis at (0, −3).
(a)
Write down the x-coordinates of the stationary points of C.
(1)
(b)
Using Figure 1, determine the nature of each stationary point of C. Give a reason for each answer.
(2)
(c)
Find an expression for f′(x) in the form x2 + px + q.
(2)
(d)
Write down the set of values of x for which f is a decreasing function.
(1)
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Answer
Mark
Notes
Part (a)
x = −1 and x = 3
B1
Part (b)
considers the sign of f′(x) either side of each value
M1
at x = −1, f′ changes from positive to negative, so a maximum; at x = 3, f′ changes from negative to positive, so a minimum
A1
Part (c)
f′(x) = a(x + 1)(x − 3) and uses (0, −3) to give a = 1
M1
f′(x) = x2 − 2x − 3
A1
Part (d)
−1 < x < 3
B1
Common error: Reading the turning point of the graph in Figure 1, at x = 1, as a stationary point of C. That point is where f′ is least; the stationary points of C are where f′(x) = 0, on the x-axis.
A farmer uses 240 m of fencing to make three identical rectangular pens, arranged side by side as shown in the plan in Figure 1. The outside of the enclosure is a rectangle measuring x m by y m, and two internal fences, each of length x m, divide it into the three pens. All 240 m of fencing is used. The total area of the enclosure is A m2.
(a)
Show that A = 120x − 2x2.
(2)
(b)
Use calculus to find the maximum total area, and the dimensions of the enclosure that give it. Justify that your area is a maximum.
(4)
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Answer
Mark
Notes
Part (a)
4x + 2y = 240, so y = 120 − 2x
M1
A = x(120 − 2x) = 120x − 2x2 as required
A1
Part (b)
dA/dx = 120 − 4x = 0
M1
x = 30, y = 60
A1
A = 1800 m2, enclosure 30 m by 60 m
A1
d2A/dx2 = −4 < 0, so a maximum
B1
Common error: Counting only two lengths of x. In Figure 1 there are four fences of length x m (two ends and two internal fences) but only two of length y m.
A rectangular sheet of card measures 40 cm by 25 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box (a box with no lid). The volume of the box is V cm3.
Figure: the card with a square of side x cm cut from each corner; fold along the dashed lines (not to scale)
(a)
Show that V = 4x3 − 130x2 + 1000x.
(2)
(b)
Explain why 0 < x < 12.5.
(1)
(c)
Find the value of x for which V is stationary, and the corresponding volume.
(4)
(d)
Show that this volume is a maximum.
(2)
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Answer
Mark
Notes
Part (a)
base (40 − 2x) by (25 − 2x), height x
M1
expands x(40 − 2x)(25 − 2x) to the given result
A1
Part (b)
x is a length so x > 0, and the width 25 − 2x must be positive, so x < 12.5
B1
Part (c)
dV/dx = 12x2 − 260x + 1000
M1
3x2 − 65x + 250 = 0, so (x − 5)(3x − 50) = 0
M1
x = 5 (rejecting 50/3 because it exceeds 12.5)
A1
V = 2250 cm3
A1
Part (d)
d2V/dx2 = 24x − 260
M1
at x = 5: −140 < 0, so maximum
A1
Exam tip: Always test each solution against the physical range. Here 50/3 would need squares wider than half the short side of the card.
A flower bed is in the shape of a sector of a circle with radius r metres and angle θ radians. The bed is edged all the way round its perimeter (the two radii and the arc) with 36 m of edging. The area of the bed is A m2. (Arc length = rθ and sector area = ½r2θ.)
Figure: the flower bed: a sector of radius r m and angle θ radians (not to scale)
(a)
Show that A = 18r − r2.
(3)
(b)
Use calculus to find the maximum area of the flower bed, justifying that it is a maximum.
(3)
(c)
Find the value of θ when the area is a maximum.
(1)
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Answer
Mark
Notes
Part (a)
2r + rθ = 36
M1
θ = (36 − 2r)/r
A1
A = ½r2 × (36 − 2r)/r = 18r − r2 as required
A1
Part (b)
dA/dr = 18 − 2r = 0, so r = 9
M1
A = 81 m2
A1
d2A/dr2 = −2 < 0, so maximum
B1
Part (c)
θ = (36 − 18)/9 = 2 radians
B1
Exam tip: Eliminate the angle first: the constraint (the perimeter) gives θ in terms of r, and only then can the area be differentiated.
For non-zero real values of x, a function is given by y = f(x) = x2 − 54/x. Its graph has exactly one turning point, which lies to the left of the y-axis.
(a)
Find the coordinates of this stationary point.
(4)
(b)
Determine the nature of the stationary point.
(2)
(c)
Show that y is increasing for all x > 0.
(2)
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Answer
Mark
Notes
Part (a)
writes 54/x as 54x−1 and differentiates
M1
dy/dx = 2x + 54x−2
A1
2x = −54/x2, so x3 = −27
M1
(−3, 27)
A1
Part (b)
d2y/dx2 = 2 − 108x−3
M1
at x = −3: 2 + 4 = 6 > 0, so a minimum
A1
Part (c)
for x > 0, 2x > 0 and 54/x2 > 0
M1
so dy/dx > 0 for all x > 0, hence y is increasing there
A1
Model sketch
Figure: model sketch of y = f(x): one stationary point, a minimum at (−3, 27)
Common error: Differentiating −54x−1 as −54x−2. Multiplying by the power −1 changes the sign: the result is +54x−2.
Figure 1 shows the curve C with equation y = 2x + 8⁄x2, x > 0. The point M is the stationary point of C. The dashed line y = k, where k is a constant, meets C at two points.
(a)
Find dy/dx.
(2)
(b)
Find the coordinates of M.
(2)
(c)
Find d2y/dx2 and use it to show that M is a minimum point.
(2)
(d)
Using Figure 1, find the set of values of k for which the line y = k meets C at two distinct points.
(2)
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Answer
Mark
Notes
Part (a)
writes 8⁄x2 as 8x−2 and differentiates, power reduced by one
M1
dy/dx = 2 − 16x−3
A1
Part (b)
2 = 16/x3, so x3 = 8
M1
M(2, 6)
A1
Part (c)
d2y/dx2 = 48x−4
M1
at x = 2: 48/16 = 3 > 0, so M is a minimum
A1
Part (d)
the line must lie above the minimum point, since C rises without limit on both sides of M
M1
k > 6 ft their M
A1
Common error: Writing k ≥ 6. When k = 6 the line touches C at M only, which is one point, not two.
Figure 1 shows part of the curve with equation y = x3 + px2 + qx + r, where p, q and r are constants. The curve has a maximum point A and a minimum point B, whose x-coordinates can be read from the grid. The curve crosses the y-axis at (0, 2).
Figure 1 shows the curve with equation y = √x, x ≥ 0, and the point A(5, 0). The point P(x, √x) lies on the curve.
(a)
Show that AP2 = x2 − 9x + 25.
(2)
(b)
Use calculus to find the value of x for which AP2 is stationary, and show that this gives a minimum.
(3)
(c)
Hence find the exact coordinates of the point on the curve that is closest to A, and the exact shortest distance from A to the curve.
(3)
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Answer
Mark
Notes
Part (a)
AP2 = (x − 5)2 + (√x)2
M1
= x2 − 10x + 25 + x = x2 − 9x + 25 as required
A1
Part (b)
d(AP2)/dx = 2x − 9 = 0
M1
x = 4.5
A1
second derivative = 2 > 0, so a minimum
B1
Part (c)
y = √4.5 = 3√2/2, so the point is (9/2, 3√2/2) ft their x
B1
AP2 = 20.25 − 40.5 + 25 = 19/4
M1
shortest distance √19/2 (≈ 2.18)
A1
Exam tip: Minimise AP2 rather than AP: the distance is never negative, so AP and AP2 are least at the same point, and AP2 has no square root to differentiate.
Figure 1 shows part of the curve with equation y = 9 − x2. The point P(x, 9 − x2) lies on the curve, with 0 < x < 3, and N is the point on the x-axis vertically below P. The shaded triangle OPN, where O is the origin, has area A.
(a)
Show that A = ½(9x − x3).
(2)
(b)
Explain why the condition 0 < x < 3 is needed.
(1)
(c)
Use calculus to find the exact value of x for which A is stationary.
(3)
(d)
Find the exact maximum area of the triangle, and justify that it is a maximum.
(2)
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Answer
Mark
Notes
Part (a)
ON = x and PN = 9 − x2
B1
A = ½x(9 − x2) = ½(9x − x3) as required
A1
Part (b)
x = 0 or x = 3 gives no triangle, and for x > 3 the point P lies below the x-axis
B1
Part (c)
dA/dx = ½(9 − 3x2)
M1
9 − 3x2 = 0, so x2 = 3
M1
x = √3 (rejecting −√3)
A1
Part (d)
A = ½√3(9 − 3) = 3√3
A1
d2A/dx2 = −3x = −3√3 < 0, so a maximum
B1
Common error: Using the height 9 for every triangle. The height PN changes as P moves along the curve: it is the y-coordinate of P, 9 − x2.
The temperature θ °C at the centre of a compost heap is monitored for 9 days. A model for θ is θ = 10 + ¼(9t2 − t3), 0 ≤ t ≤ 9, where t is the time in days. Figure 1 shows the graph of θ against t given by the model.
(a)
Use Figure 1 to estimate the greatest temperature predicted by the model and the time at which it occurs.
(2)
(b)
Use calculus to confirm the time at which the temperature is greatest, find the exact greatest temperature, and justify that it is a maximum.
(4)
(c)
Find the time at which the temperature is rising most rapidly, and the rate of increase at that time.
(3)
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Answer
Mark
Notes
Part (a)
t = 6 (allow 5.5 to 6.5)
B1
θ ≈ 37 (allow 36 to 38)
B1
Part (b)
dθ/dt = ¼(18t − 3t2)
B1
¾t(6 − t) = 0, so t = 6 (t = 0 is the start of the model)
M1
θ = 37 °C
A1
d2θ/dt2 = ¼(18 − 6t) = −4.5 < 0 at t = 6, so a maximum
B1
Part (c)
sets d2θ/dt2 = 0, i.e. maximises dθ/dt
M1
t = 3 days
A1
rate = ¼(54 − 27) = 6.75 °C per day
A1
Exam tip: “Rising most rapidly” asks for the greatest value of the gradient dθ/dt, so differentiate the gradient and set that equal to zero. On the graph this is where the curve is steepest, not where it is highest.
Figure 1 shows an open-topped box (with no lid) made from thin card. The box has a square base of side x cm and height h cm. A vertical partition of card, also x cm by h cm, divides the box into two equal compartments. The total area of card used for the base, the four sides and the partition is 1200 cm2. The volume of the box is V cm3.
(a)
Show that V = 240x − 1⁄5x3.
(3)
(b)
Use calculus to find the value of x for which V is stationary, and find this value of V.
(4)
(c)
Justify that this value of V is a maximum, and find the height of the box in this case.
(2)
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Answer
Mark
Notes
Part (a)
x2 + 4xh + xh = 1200
M1
h = (1200 − x2)/(5x)
A1
V = x2h = x(1200 − x2)/5 = 240x − 1⁄5x3 as required
A1
Part (b)
dV/dx = 240 − 3⁄5x2
M1
x2 = 400
M1
x = 20
A1
V = 3200 cm3
A1
Part (c)
d2V/dx2 = −6⁄5x = −24 < 0, so a maximum
B1
h = 8 cm
B1
Common error: Leaving out the partition, or including a lid. Count the card piece by piece from Figure 1: one base, four sides and one partition, which is x2 + 5xh.
A closed box is a cuboid with a rectangular base of length 2x cm and width x cm, and height h cm. The volume of the box is 576 cm3. The total surface area of the box is S cm2.
Figure: the closed box (not to scale)
(a)
Show that S = 4x2 + 1728/x.
(4)
(b)
Use calculus to find the value of x for which S is stationary, and the corresponding value of S.
(4)
(c)
Justify that this value of S is a minimum.
(2)
(d)
State the dimensions of the box with the least surface area.
(1)
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Answer
Mark
Notes
Part (a)
2x2h = 576, so h = 288/x2
M1
S = 2(2x × x) + 2(2xh) + 2(xh)
M1
= 4x2 + 6xh
A1
= 4x2 + 6x × 288/x2 = 4x2 + 1728/x as required
A1
Part (b)
differentiates, power of 1728x−1 reduced
M1
dS/dx = 8x − 1728/x2
A1
x3 = 216, so x = 6
A1
S = 432 cm2
A1
Part (c)
d2S/dx2 = 8 + 3456/x3
M1
at x = 6: 8 + 16 = 24 > 0, so minimum
A1
Part (d)
12 cm by 6 cm by 8 cm
B1
Exam tip: Count the faces carefully before writing S: a closed cuboid has three pairs of equal faces, and each pair contributes twice its area.
An open-topped water tank is a cylinder of radius r metres and height h metres, with a base but no lid. It is made from 108π m2 of sheet metal, all of which is used for the base and the curved side. The volume of the tank is V m3.
(a)
Show that V = 54πr − ½πr3.
(4)
(b)
Use calculus to find the value of r for which V is stationary, and find this value of V, giving your answer as an exact multiple of π.
(4)
(c)
Justify that this value of V is a maximum.
(2)
(d)
Show that, when V is a maximum, the height of the tank equals its radius.
(1)
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Answer
Mark
Notes
Part (a)
πr2 + 2πrh = 108π
M1
h = (108 − r2)/(2r)
A1
V = πr2h with their h substituted
M1
= πr(108 − r2)/2 = 54πr − ½πr3 as required
A1
Part (b)
dV/dr = 54π − (3/2)πr2
M1
= 0 gives r2 = 36, so r = 6 (rejecting −6)
A1
substitutes r = 6: 324π − 108π
M1
V = 216π m3
A1
Part (c)
d2V/dr2 = −3πr
M1
at r = 6: −18π < 0, so maximum
A1
Part (d)
h = (108 − 36)/12 = 6 = r
B1
Common error: Using the surface area of a closed cylinder. With no lid there is only one circular end: πr2 + 2πrh.
A rectangular yard of area 600 m2 is built against the straight wall of a warehouse, which forms one side and needs no fencing. The side opposite the wall has length x metres and is fenced with panels costing £24 per metre. The two sides perpendicular to the wall are fenced with panels costing £16 per metre. The total cost of the fencing is £C.
Figure: the yard against the wall (not to scale)
(a)
Show that C = 24x + 19200/x.
(3)
(b)
Use calculus to find the exact value of x for which C is stationary.
(4)
(c)
Justify that this value of x gives the minimum cost.
(2)
(d)
Find the minimum cost, giving your answer in the form p√2, and also to the nearest pound.
(2)
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Answer
Mark
Notes
Part (a)
perpendicular sides each 600/x m
B1
C = 24x + 2 × 16 × 600/x
M1
= 24x + 19200/x as required
A1
Part (b)
differentiates 19200x−1 with power reduced
M1
dC/dx = 24 − 19200/x2
A1
x2 = 800
M1
x = 20√2
A1
Part (c)
d2C/dx2 = 38400/x3
M1
positive for all x > 0, so a minimum
A1
Part (d)
24 × 20√2 + 19200/(20√2) = 480√2 + 480√2
M1
£960√2 ≈ £1358
A1
Exam tip: Keep the surd until the last line: 19200/(20√2) = 960/√2 = 480√2, so the two parts of the cost are equal at the minimum.
A closed container is a prism of length L cm whose cross-section is an equilateral triangle of side x cm, as shown in Figure 1. The volume of the container is 54 cm3. The total surface area of the container, including both triangular ends, is S cm2.
(a)
Show that the area of one triangular end is √3⁄4x2.
(2)
(b)
Show that S = √3⁄2x2 + 216√3⁄x.
(4)
(c)
Use calculus to find the value of x for which S is stationary.
(3)
(d)
Find the exact minimum surface area and the corresponding length of the container.
(2)
(e)
Justify that this value of S is a minimum.
(2)
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Answer
Mark
Notes
Part (a)
height of the triangle = x sin 60° = √3⁄2x (or by Pythagoras)
M1
area = ½ × x × √3⁄2x = √3⁄4x2 as required
A1
Part (b)
√3⁄4x2L = 54
M1
L = 216/(√3x2) = 72√3/x2
A1
S = 2 × √3⁄4x2 + 3xL
M1
= √3⁄2x2 + 216√3/x as required
A1
Part (c)
dS/dx = √3x − 216√3x−2
M1
√3x = 216√3/x2, so x3 = 216
M1
x = 6
A1
Part (d)
S = 18√3 + 36√3 = 54√3 cm2 (≈ 93.5)
A1
L = 2√3 cm
A1
Part (e)
d2S/dx2 = √3 + 432√3x−3
M1
at x = 6: √3 + 2√3 = 3√3 > 0, so a minimum
A1
Exam tip: Count the faces from Figure 1 before writing S: two equilateral triangles and three identical rectangles, each x by L. Keep √3 as a common factor so the stationary-point equation stays simple.
A solid sphere has centre O and radius 6√3 cm. A cylinder of radius r cm and height h cm is cut from the sphere, with the axis of the cylinder passing through O and both circular edges of the cylinder lying on the surface of the sphere. Figure 1 shows a cross-section through O. The volume of the cylinder is V cm3. (The volume of a sphere of radius R is 4⁄3πR3.)
(a)
Show that V = 108πh − ¼πh3.
(3)
(b)
Use calculus to find the value of h for which V is stationary.
(3)
(c)
Find the exact maximum volume of the cylinder, and the exact radius of this cylinder.
(2)
(d)
Justify that this volume is a maximum.
(2)
(e)
Show that the largest cylinder occupies a fraction 1⁄√3 of the volume of the sphere, and give this as a percentage to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
Pythagoras in the cross-section: r2 + (h/2)2 = (6√3)2
M1
r2 = 108 − ¼h2
A1
V = πr2h = π(108 − ¼h2)h = 108πh − ¼πh3 as required
Common error: Using h instead of h/2 in Pythagoras. The right-angled triangle in Figure 1 runs from O to a corner of the rectangle, so its vertical side is half the height of the cylinder.
A decorative glass panel is made in the shape shown in Figure 1: a rectangle of width x m and height h m with an equilateral triangle of side x m on its top edge. A metal strip of total length 6 m goes all the way round the outside edge of the panel (the heavy line); there is no strip along the dashed line. The area of the panel is A m2.
(a)
Show that A = 3x − 6 − √3⁄4x2.
(4)
(b)
Use calculus to find the value of x for which A is stationary, giving your answer in the form a + b√3⁄c, where a, b and c are integers.
(4)
(c)
Justify that this value of x gives the maximum area.
(1)
(d)
Find the maximum area, giving your answer to 3 significant figures.
(2)
(e)
Show that, for the panel of maximum area, the height h is less than the width x.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
3x + 2h = 6
M1
h = 3 − 3⁄2x
A1
area of triangle = √3⁄4x2
B1
A = x(3 − 3⁄2x) + √3⁄4x2 = 3x − 6 − √3⁄4x2 as required
A1
Part (b)
dA/dx = 3 − 6 − √3⁄2x
M1
sets equal to 0 and solves
M1
x = 6/(6 − √3)
A1
rationalises: 6(6 + √3)/33 = (12 + 2√3)/11
A1
Part (c)
d2A/dx2 = −6 − √3⁄2 < 0 because 6 > √3, so a maximum
B1
Part (d)
substitutes x ≈ 1.4058 into A
M1
2.11 m2
A1
Part (e)
h = 3 − 1.5 × 1.4058 ≈ 0.891 m, which is less than x ≈ 1.41 m
B1
Exam tip: The strip runs only round the outside, so the top edge of the rectangle is not included: the perimeter is x + 2h + 2x, not 2x + 2h + 2x.
The curve C has equation y = f(x), where f(x) = x4 − 6x2 + 8x + 5. Figure 1 shows part of C. The points A and B are the stationary points of C.
(a)
Show that f′(x) = 4(x − 1)2(x + 2).
(3)
(b)
Find the coordinates of A and of B.
(2)
(c)
Use the second derivative to show that A is a minimum point.
(2)
(d)
Show that the second derivative test fails at B, and determine the nature of B.
(3)
(e)
Find the set of values of x for which f is decreasing.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
f′(x) = 4x3 − 12x + 8
B1
expands 4(x2 − 2x + 1)(x + 2) or uses the factor theorem
M1
= 4(x3 − 3x + 2), matching f′(x)
A1
Part (b)
x = −2 and x = 1 from f′(x) = 0
M1
A(−2, −19), B(1, 8)
A1
Part (c)
f″(x) = 12x2 − 12
M1
at x = −2: 36 > 0, so A is a minimum
A1
Part (d)
at x = 1: f″ = 12 − 12 = 0, so the test gives no conclusion
B1
gradient either side, e.g. at x = 0.5: 4(0.25)(2.5) = 2.5, and at x = 1.5: 4(0.25)(3.5) = 3.5
M1
positive on both sides, so B is a stationary point of inflection
A1
Part (e)
f′(x) < 0 only when x + 2 < 0: x < −2
B1
Exam tip: In factorised form the sign of f′ is easy to read: (x − 1)2 is never negative, so the sign of f′(x) is the sign of x + 2. That explains both the minimum at A and the inflection at B.
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