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Topic 5.09 · HL only

Second derivative (+max/min distinguishing): notes and practice questions

Summary
  • Second order derivative: f′′(x)f''(x) or d2ydx2\frac{d^2y}{dx^2}, obtained by differentiating the first derivative f′(x)f'(x).
  • Local minimum/maximum points: Stationary points where the gradient f′(x)=0f'(x) = 0.
  • Concave up: Curve forms a "U" shape, gradient is increasing, meaning f′′(x)>0f''(x) > 0.
  • Concave down: Curve forms an "n" shape, gradient is decreasing, meaning f′′(x)<0f''(x) < 0.
  • Point of Inflection: A point where the curve changes concavity.
  • Second Derivative Test for Stationary Points:
  • If f′′(x)>0f''(x) > 0, the stationary point is a local minimum.
  • If f′′(x)<0f''(x) < 0, the stationary point is a local maximum.
  • If f′′(x)=0f''(x) = 0, the test is inconclusive; further investigation (e.g., first derivative test or sign change of f′′(x)f''(x)) is required.
  • Conditions for a Point of Inflection:

1. The second derivative is zero: f′′(x)=0f''(x) = 0.
2. The second derivative f′′(x)f''(x) must change sign (concavity changes) as it passes through the point.

  • To find and classify stationary points algebraically:

1. Find f′(x)f'(x).
2. Set f′(x)=0f'(x) = 0 and solve for x-coordinates of stationary points.
3. Substitute x-values into the original function f(x)f(x) to find y-coordinates.
4. Find f′′(x)f''(x).
5. Substitute x-coordinates into f′′(x)f''(x): positive result indicates local minimum, negative result indicates local maximum, zero result is inconclusive.

  • GDC can find turning points and visually classify them; use to check algebraic solutions.
  • Second derivative, concavity, and points of inflection are HL only topics.
  • "Classify turning points" means to determine their nature (local min/max).
  • Exercise caution with negative and fractional powers during differentiation.
  • For a point of inflection, f′′(x)=0f''(x) = 0 is necessary but not sufficient; a change in concavity (sign change of f′′(x)f''(x)) must also occur.

How it is examined

The second derivative test is the HL-only shortcut: compute f′′(x)f''(x) at the stationary point, positive means a minimum, negative means a maximum, zero means the test is inconclusive and the student has to fall back on the SL method (checking the sign of f′(x)f'(x) either side). A point of inflexion question wants the concavity language explicitly, "concave-up" or "concave-down", not just "the curve changes shape."

Key ideas
  • Find the second derivative, in both notations d2ydx2\dfrac{d^2y}{dx^2} and f′′(x)f''(x).
  • Use the second derivative test to distinguish a maximum from a minimum point.
  • Recognise that a point of inflexion is where concavity changes, and interpret this in context.
  • Use "concave-up" for f′′(x)>0f''(x) > 0 and "concave-down" for f′′(x)<0f''(x) < 0.

Linking questions

  • Links to other subjects: simple harmonic motion (physics).
  • TOK: music can be expressed using mathematics. Does that mean music is mathematical, or mathematics musical?

Practice questions

15 questions · 12 medium · 3 hard
Showing 15 of 15

Question 1

MediumPaper 1 · calculator6 marks
(a)

Consider the function y=(x3−2x2)exy = \left( x^{3} - 2x^{2} \right)e^{x}.

(a) Find d2ydx2\frac{d^{2}y}{dx^{2}}.

[4]
(b)

(b) Find how many points of inflection does this function contain.

[2]

Question 2

HardPaper 1 · calculator10 marks
(a)

A group of engineers is designing a new observation Ferris wheel. The height, H(t)H(t), in metres, of a passenger capsule above the ground is modelled by the function H(t)=pcos⁡(π100t)+qH(t) = p \cos\left(\frac{\pi}{100}t\right) + q, where tt is the time in seconds after the capsule begins its ascent from the highest point.

The lowest point a capsule reaches is 2 metres above the ground, and the highest point is 32 metres above the ground. The Ferris wheel completes one full rotation in 200 seconds.

Find the values of pp and qq.

[2]
(b)

Using your values from part (a), the function is H(t)=15cos⁡(π100t)+17H(t) = 15 \cos\left(\frac{\pi}{100}t\right) + 17.

(i) Find H′(t)H'(t).

(ii) Find H′′(t)H''(t).

[3]
(c)(i)

The engineers are particularly interested in the moment when the capsule's vertical speed is at its maximum, for the first time after t=0t=0. This occurs at time t=kt=k.

Calculate the value of kk.

[3]
(c)(ii)

Calculate the height of the capsule at this time kk.

[2]

Question 3

MediumPaper 1 · calculator6 marks
(a)

Consider the function y=(x3−2x2)exy = \left( x^{3} - 2x^{2} \right)e^{x}.

aa Find d2ydx2\frac{d^{2}y}{dx^{2}}.

[4]
(b)

bb Find how many points of inflection does this function contain.

[2]

Question 4

HardPaper 2 · calculator15 marks
(a)

(a) A pharmaceutical company is designing a new cylindrical container for a special liquid. The container has a fixed volume V=500 cm3V = 500 \text{ cm}^3.

The radius of the container is r cmr \text{ cm} and the height is h cmh \text{ cm}.

Show that πr2h=500\pi r^2 h = 500.

[2]
(b)

(b) Find an expression for the total surface area SS of the container.

[2]
(c)

(c) Substitute an expression for hh (from part (a) ) into your expression for SS (from part (b) ) and hence show that S=2πr2+1000rS = 2\pi r^2 + \frac{1000}{r}.

[3]
(d)

(d) Find dSdr\frac{dS}{dr}.

[2]
(e)

(e) Find the minimum value of SS and the values of rr and hh when this occurs. Show that this value of SS is indeed a minimum.

[6]

Question 5

MediumPaper 1 · calculator5 marks
(a)

The diagram shows the slope field for the differential equation dydx=cos⁡(x−y) \frac{dy}{dx} = \cos(x-y) for −π≤x≤π -\pi \le x \le \pi and −π≤y≤π -\pi \le y \le \pi .

Slope field for dy/dx = cos(x-y) with two solution curves and lines L1 and L2.

The local maximum points for solutions to the differential equation lie on the straight line L1 L_1 .

Find the equation of L1 L_1 , giving your answer in the form y=mx+c y = mx + c .

[3]
(b)

Find the equation of the straight line L2 L_2 on which all local minimum points lie within the given domain, giving your answer in the form y=mx+c y = mx + c .

[2]

Question 6

HardPaper 2 · calculator20 marks
(a)

(a) SweetTreats is designing new packaging for a line of artisanal chocolates. The initial design is a box in the shape of a cuboid with a square base of side length LL cm. Its height, HH cm, is twice the length of the base.

Write down an expression for HH in terms of LL.

[1]
(b)

(b) The box is designed to hold 250250 cm3^3 of chocolates.

Find the value of LL and HH.

[3]
(c)

(c) Calculate the total external surface area of the box.

[3]
(d)

(d) To minimize the amount of material needed, SweetTreats is considering changing the shape to a cylinder with radius rr cm and height hh cm. The cylindrical container must also hold 250250 cm3^3 of chocolates.

Find an expression for the height, hh, of the container in terms of rr.

[2]
(e)

(e) Let the total external surface area of the cylindrical container be AA cm2^2.

Show that A=2πr2+500rA = 2\pi r^2 + \frac{500}{r}.

[2]
(f)

(f) Find dAdr\frac{\text{d}A}{\text{d}r}.

[3]
(g)(i)

(g.i) Hence or otherwise, find the value of rr that will minimize AA.

[2]
(g)(ii)

(g.ii) Find the minimum value of AA needed for the cylinder.

[1]
(h)(i)

(h.i) Find d2Adr2\frac{\text{d}^2 A}{\text{d}r^2}.

[1]
(h)(ii)

(h.ii) Hence determine whether the graph of AA is concave-up or concave-down for r>0r > 0. Justify your answer.

[2]

Question 7

MediumPaper 1 · calculator10 marks
(a)(i)

A company models the profit from a new product launch using the function P(t)=3t(2−e−t)P(t) = 3t(2 - e^{-t}), where PP is the profit in thousands of dollars and tt is the time in months since launch.

Find dPdt\frac{dP}{dt}.

[4]
(a)(ii)

Find d2Pdt2\frac{d^2P}{dt^2}.

[4]
(b)

The company observes that the rate of change of profit growth begins to slow down after a certain point, indicating a point of inflexion. Find the time, tt, in months at which this point of inflexion occurs.

[2]

Question 8

MediumPaper 1 · calculator10 marks
(a)

(a) A drone's displacement, ss metres, from a fixed charging station after tt minutes (where t≥0t \geq 0) is given by the equation s(t)=t2t+1+ln⁡(t+1)s(t) = \frac{t^2}{t+1} + \ln(t+1).

Calculate the distance travelled by the drone in the first 3 minutes.

[4]
(b)

(b) Find an expression for the velocity, v(t)v(t), of the drone. Hence, determine if the drone ever becomes stationary for t≥0t \geq 0.

[6]

Question 9

MediumPaper 1 · calculator14 marks
(a)

(a) A company's profit, PP, in thousands of dollars, from selling xx units of a new product can be modelled by the function P(x)=x3−3x2−9x+5P(x) = x^3 - 3x^2 - 9x + 5, for x≥0x \ge 0.

Calculate the coordinates of the stationary points of the profit function P(x)P(x).

[4]
(b)

(b) Find the equation of the normal to the curve y=P(x)y = P(x) at the point where x=1x = 1. Give your answer in the form ax+by+d=0ax + by + d = 0, where a,b,d∈Za, b, d \in \mathbb{Z}.

[5]
(c)

(c) Determine the second derivative of the profit function, P′′(x)P''(x).

[2]
(d)

(d) Use the second derivative to classify the nature of the stationary points found in part (a).

[3]

Question 10

MediumPaper 2 · calculator13 marks
(a)

(a) A builder is designing a rectangular enclosure for a new community garden. The total length of fencing available for the enclosure is 6060 metres.

The builder wants to maximize the area of the garden. Calculate the maximum possible area of the enclosure.

[5]
(b)

(b) To ensure structural stability, the builder also needs to consider the sum of the squares of the lengths of the sides. Calculate the minimum possible value for the sum of the squares of the lengths of the sides of the enclosure.

[5]
(c)

(c) Justify that the value found in part (b) is indeed a minimum.

[3]

Question 11

MediumPaper 1 · calculator7 marks
(a)

(a) The profit, PP, in thousands of dollars, for a company producing xx hundred units of a certain product is modelled by the function P(x)=−2x2+20x−15P(x) = -2x^2 + 20x - 15, for x≥0x \ge 0.

Find an expression for dPdx\frac{dP}{dx}, the rate of change of profit with respect to the number of units produced.

[2]
(b)

(b) Hence, find the maximum profit the company can achieve.

[3]
(c)

(c) Justify that the profit you found in part (b) is a maximum point.

[2]

Question 12

MediumPaper 2 · calculator16 marks
(a)

A civil engineer is designing a cross-section for a new pedestrian tunnel. The shape of the tunnel's ceiling can be modelled by the curve y=10−x24y = 10 - \frac{x^2}{4}, where xx and yy are measured in metres. The tunnel's base is along the xx-axis, and it starts from the yy-axis, extending to where the ceiling meets the xx-axis.

(a) Sketch the curve, showing the region representing the cross-sectional area of the tunnel. Shade this region.

[3]
(b)

(b) Using the trapezium rule with five strips, determine an approximation for the cross-sectional area of the tunnel.

[5]
(c)

(c) Explain why your answer to part (b) will be an underestimate for the actual cross-sectional area.

[2]
(d)

(d) Using integration, determine the exact value of the cross-sectional area of the tunnel.

[4]
(e)

(e) Find the percentage error of your approximation from part (b) compared with the exact value found in part (d).

[2]

Question 13

MediumPaper 1 · calculator10 marks
(a)

(a) The height of a drone above the ground, in metres, can be modelled by the function H(t)=0.5t2−10t+70H(t) = 0.5t^2 - 10t + 70, where tt is the time in seconds after launch.

Find H′(t)H'(t).

[2]
(b)

(b) Find H′(12)H'(12) and H′(32)H'(32) and comment on the meaning of these values in the context of the drone's flight.

[4]
(c)

(c) Find the time when the drone reaches its minimum height and explain how you know that it is a minimum.

[4]

Question 14

MediumPaper 1 · calculator18 marks
(a)

The diagram below shows a cross-section of a sculpted garden feature. The shaded area is bounded by the xx-axis, the vertical line x=6x = 6, the line y=x−1y = x-1 and the curve y=18x2y = \frac{18}{x^2}.

Graph showing a shaded area bounded by x-axis, x=6, y=x-1 and y=18/x^2, with points P, Q, R, S labeled.

Using technology or otherwise, find the coordinates of points PP, QQ, RR and SS.

[4]
(b)

Approximate the area of the cross-section using the trapezoidal rule with 5 strips of equal width over the interval [1,6][1,6].

[6]
(c)

Explain why the approximation in part (b) is an overestimate.

[2]
(d)

Determine the exact area of the cross-section using integration.

[4]
(e)

Find the percentage error of your approximation from part (b), compared with the exact value from part (d).

[2]

Question 15

MediumPaper 1 · calculator5 marks
(a)

The rate of a chemical reaction, RR, in mol/s, at time tt seconds, is modelled by the function R(t)=10t2−3t+5R(t) = \frac{10}{t^2 - 3t + 5}, for t≥0t \ge 0.

(a) Find an expression for R′(t)R'(t).

[2]
(b)

(b) Hence or otherwise, find the values of tt at which the rate of reaction has points of inflexion. Give your answers correct to three significant figures.

[3]

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What does Second derivative (+max/min distinguishing) cover in IB Maths AI?

Second order derivative: f''(x) or (d^2y)/(dx^2), obtained by differentiating the first derivative f'(x). Local minimum/maximum points: Stationary points where the gradient f'(x) = 0. Concave up: Curve forms a "U" shape, gradient is increasing, meaning f''(x) > 0.

Is Second derivative (+max/min distinguishing) SL or HL?

Second derivative (+max/min distinguishing) is HL only. SL students are not examined on it.

How do I revise Second derivative (+max/min distinguishing) for IB Maths AI?

Start from the core idea: second order derivative: f''(x) or (d^2y)/(dx^2), obtained by differentiating the first derivative f'(x). In the exam: the second derivative test is the HL-only shortcut: compute f''(x) at the stationary point, positive means a minimum, negative means a maximum, zero means the test is inconclusive and the student has to fall back on the SL method (checking the sign of f'(x) either side). A point of inflexion question wants the concavity language explicitly, "concave-up" or "concave-down", not just "the curve changes shape.". Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Second derivative (+max/min distinguishing)?

FourtyFive has 15 Second derivative (+max/min distinguishing) questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

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