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Topic 5.06 · SL and HL

Second derivative, points of inflection, testing max / min points, optimisation: notes and practice questions

Summary
  • Points of Inflection: Occur when f′′(x)=0f''(x) = 0 or undefined, and the concavity changes (verify by checking sign changes in f′′(x)f''(x)).
  • Testing Max/Min Points:
  • f′(x)=0f'(x) = 0 identifies critical points.
  • f′′(x)>0f''(x) > 0: Minimum, f′′(x)<0f''(x) < 0: Maximum.
  • Optimisation: Involves finding maximum or minimum values of functions in real-world contexts.

How it is examined

"Here is the graph of f′f', sketch ff" and its variants are the reasoning question and they are hard for students. Zeros of f′f' are stationary points of ff, zeros of f′′f'' with a sign change are inflexion points of ff. 3 to 5 marks. The y=x4y = x^4 counterexample is explicit syllabus content, so a question can ask why f′′(x)=0f''(x) = 0 is not enough, and the answer must mention the sign change. Optimization questions want the constraint used to reduce to one variable, then a justification that the stationary point is the maximum, and that justification is a separate mark. 6 to 9 marks across parts.

Key ideas
  • The second derivative.
  • Graphical behaviour of functions, including the relationship between the graphs of ff, f′f' and f′′f''.
  • Local maximum and minimum points.
  • Testing for maximum and minimum.
At HL
  • Extended at AHL 5.12 (higher derivatives, f(n)(x)f^{(n)}(x)).
  • Extended at AHL 5.14 (optimisation including cases where the optimum is at an end point).

Linking questions

  • Feeds SL 5.8 immediately; in practice they are taught and examined together.
  • Other contexts: profit, area, volume.
  • Links to other subjects: allocative efficiency (economics).

Practice questions

65 questions · 1 easy · 33 medium · 31 hard
Showing 20 of 20

Question 1

EasyPaper 1 · no calculator5 marks

Consider the function f(x)=5xe−xf(x) = 5xe^{- x}, where xϵRx\epsilon\mathbb{R}. At x=ax = a there is a point of inflection. Find the value of aa.

Question 2

MediumPaper 1 · no calculator8 marks
(a)

Consider the function ff, which has the derivative f′(x)=−4x2+28x+c,f^{'}(x) = - 4x^{2} + 28x + c, where c∈Zc\mathbb{\in Z}.

aa Knowing that the y coordinate of the vertex of f′f^{'} is equal to 9, find the value of cc.

[4]
(b)

bb For what values of xx is the graph of ff concaved upwards. Justify your answer.

[4]

Question 3

HardPaper 3 · calculator24 marks
(a)

A biologist is modelling the growth of two different bacterial colonies. The first colony, A, grows such that its population at time xx is given by PA(x)=axP_A(x) = a^x, where aa is a growth factor and x≥0x \ge 0. The second colony, B, grows linearly such that its population at time xx is PB(x)=xP_B(x) = x.

Consider the cases where the growth factor a=2a = 2 and a=10a = 10. On the same set of axes, sketch the following three graphs for x≥0x \ge 0:

y=2xy = 2^x

y=10xy = 10^x

y=xy = x

Clearly label each graph with its equation and state the coordinates of any non-zero yy-axis intercepts.

[4]
(b)

In parts (b) and (c), consider the case where the growth factor a=ea = e.

Use calculus to find the minimum value of the expression ex−xe^x - x, justifying that this value is a minimum.

[5]
(c)

Hence deduce that ex>xe^x > x for all x∈Rx \in \mathbb{R}.

[1]
(d)

There exist values of aa for which the graph of y=axy = a^x and the line y=xy = x have different numbers of intersection points. The following table gives three intervals for the value of aa.

IntervalNumber of intersection points
0<a<10 < a < 1pp
1<a<1.41 < a < 1.4qq
1.5<a<21.5 < a < 2rr

By investigating the graph of y=axy = a^x for different values of aa, write down the values of p,qp, q and rr.

[4]
(e)

In parts (e) and (f), consider a∈R+,a≠1a \in \mathbb{R}^+, a \neq 1.

For 1.4≤a≤1.51.4 \leq a \leq 1.5, a value of aa exists such that the line y=xy = x is a tangent to the graph of y=axy = a^x at a point P.

Find the exact coordinates of P and the exact value of aa.

[8]
(f)(i)

Write down the exact set of values for aa such that the graphs of y=axy = a^x and y=xy = x have

(i) two intersection points;

[1]
(f)(ii)

(ii) no intersection points.

[1]

Question 4

MediumPaper 1 · no calculator13 marks
(a)

A function, gg, has its derivative given by g′(x)=−2x2+8x+kg'(x) = -2x^2 + 8x + k, where k∈Rk \in \mathbb{R}. The following diagram shows part of the graph of g′g'.

Graph of g' showing a parabola opening downwards, with vertex in the first quadrant.

The graph of g′g' has an axis of symmetry x=qx = q.

(a) Find the value of qq.

[2]
(b)

(b) The vertex of the graph of g′g' has a y-coordinate of 10. Find the value of kk.

[3]
(c)

(c) Find the equation of the tangent to the graph of g′g' at x=0x = 0.

[4]
(d)(i)

The graph of gg has a point of inflexion at x=cx = c.

(d) (i) Find the value of cc.

[2]
(d)(ii)

(ii) Find the values of xx for which the graph of gg is concave-up. Justify your answer.

[2]

Question 5

HardPaper 1 · no calculator19 marks
(a)

A ladder must be placed against a tall vertical building, clearing a monument that is 8 m high and stands on horizontal ground 1 m away from the building's base. The ladder touches the ground, the top corner of the monument, and the wall of the building.

A diagram showing a vertical building and the horizontal ground. A monument of height 8m stands 1m away from the base of the building. A ladder is shown leaning against the building, just touching the top of the monument. The angle the ladder makes with the ground is labelled as theta.

Let LL be the length of the ladder in metres.

Let θ\theta be the angle that the ladder makes with the ground, where 0<θ<π20 < \theta < \frac{\pi}{2}.

(a) Show that L=sec⁡θ+8csc⁡θL = \sec \theta + 8\csc \theta.

[2]
(b)(i)

(b) (i) Find dLdθ\frac{dL}{d\theta}.

[2]
(b)(ii)

(b) (ii) When dLdθ=0\frac{dL}{d\theta} = 0, show that tan⁡θ=2\tan\theta = 2.

[3]
(c)(i)

(c) (i) Find d2Ldθ2\frac{d^2L}{d\theta^2}.

[3]
(c)(ii)

(c) (ii) When tan⁡θ=2\tan\theta = 2, find the value of d2Ldθ2\frac{d^2L}{d\theta^2}.

[4]
(d)(i)

(d) (i) Hence, justify that LL is a minimum when tan⁡θ=2\tan\theta = 2.

[1]
(d)(ii)

(d) (ii) Determine this minimum value of LL.

[2]
(e)

(e) A construction company only has ladders with a maximum length of 11 m. Determine whether it is possible to position a ladder against the building over the monument, giving a reason for your answer.

[2]

Question 6

MediumPaper 1 · no calculator15 marks
(a)

A function ff is defined by f(x)=ln⁡(x)xf(x) = \frac{\ln(x)}{x}, for x>0x > 0.

The following diagram shows part of the graph of ff.

Graph of f(x) = (ln(x) )/x with an x-intercept, a local maximum M, and a point of inflection P

(a) Find the coordinates of the x-intercept of the graph of ff.

[2]
(b)

(b) Find f′(x)f'(x).

[3]
(c)

The graph of ff has a local maximum at point M.

(c) Hence, find the exact coordinates of M.

[4]
(d)(i)

(d) (i) Show that f′′(x)=2ln⁡(x)−3x3f''(x) = \frac{2\ln(x) - 3}{x^3}.

[3]
(d)(ii)

The graph of ff has a point of inflection at point P.

(d) (ii) Hence, find the exact coordinates of P.

[3]

Question 7

HardPaper 1 · no calculator14 marks
(a)

A rectangle is inscribed in an ellipse with equation x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1. The sides of the rectangle are parallel to the coordinate axes. The vertices of the rectangle are located at (±x,±y)(\pm x, \pm y), where x>0x > 0 and y>0y > 0.

Diagram of an ellipse with an inscribed rectangle

(a) Show that the area of the rectangle, AA, can be expressed as A=12x525−x2A = \frac{12x}{5}\sqrt{25-x^2}.

[4]
(b)

(b) Show that dAdx=12(25−2x2)525−x2\frac{dA}{dx} = \frac{12(25-2x^2)}{5\sqrt{25-x^2}}.

[4]
(c)

(c) Hence, find the exact dimensions of the rectangle with the maximum possible area.

[6]

Question 8

MediumPaper 1 · no calculator5 marks

Consider the function g(x)=14x4−2x2+5g(x) = \frac{1}{4}x^4 - 2x^2 + 5, where x∈Rx \in \mathbb{R}.

The graph of y=g(x)y = g(x) has a local minimum point at (p,q)(p, q) where p<0p < 0.

Find the value of pp and the value of qq.

Question 9

HardPaper 1 · no calculator15 marks
(a)

A drone takes off from a platform. Its height, hh metres, above the platform after tt seconds is given by h(t)=6t−t2h(t) = 6t - t^2, for 0≤t≤80 \le t \le 8. This is shown in the following diagram.

Graph of height h versus time t for the drone, showing a parabola opening downwards with vertex in the first quadrant and passing through the origin

The drone lands back on the platform when t=pt=p.

Find the value of pp.

[2]
(b)(i)

The drone reaches its maximum height when t=qt=q.

Find the value of qq.

[3]
(b)(ii)

Find the drone's maximum height above the platform.

[2]
(c)

Find the drone's vertical distance from the platform when t=8t=8.

[2]
(d)

The total vertical distance travelled by the drone in the first 8 seconds is given by dd.

Find the value of dd.

[2]
(e)

A second drone, Drone B, takes off from the same platform. Its velocity is given by vB(t)=8−2tv_B(t) = 8 - 2t, for t≥0t \ge 0.

When t=kt = k, the total vertical distance travelled by Drone B is equal to dd.

Find the value of kk.

[4]

Question 10

MediumPaper 2 · calculator5 marks
(a)

A pharmaceutical company is testing a new drug. The concentration of the drug in a patient's bloodstream, CC, in mg/L, tt hours after administration, is modeled by the function C(t)=10te−0.5tC(t) = 10t \text{e}^{-0.5t}, for 0≤t≤100 \le t \le 10.

Sketch the graph of C(t)C(t) on the grid below.

Graph grid with t and C(t) axes. t-axis from 0 to 10, C(t) -axis from 0 to 8.
[3]
(b)

(b) Find the time, in hours, at which the concentration of the drug in the bloodstream is at its maximum.

[2]

Question 11

HardPaper 1 · no calculator17 marks
(a)

A hollow pipe is manufactured by removing a smaller cylinder of radius rr from the centre of a larger cylinder of radius 3r3r. Both cylinders have the same height, hh. This is shown in the following diagram.

All lengths are measured in centimetres.

Diagram of a hollow cylinder with no top with outer radius 3r, inner radius r, and both with height h

The total surface area of the hollow pipe, in cm2^2, is given by SS.

(a) Show that S=16πr2+8πrhS = 16\pi r^2 + 8\pi rh.

[3]
(b)

The total surface area of the hollow pipe is 128π cm2128\pi \text{ cm}^2.

(b) Show that the volume of the pipe, VV, is given by V=128πr−16πr3V = 128\pi r - 16\pi r^3.

[6]
(c)

(c) Find an expression for dVdr\frac{dV}{dr}.

[2]
(d)

(d) The hollow pipe has its maximum volume when r=k6r = k\sqrt{6}, where k∈Q+k \in \mathbb{Q}^+. Find the value of kk.

[3]
(e)

(e) Hence, find this maximum volume, giving your answer in the form mπ6m\pi\sqrt{6}, where m∈Q+m \in \mathbb{Q}^+.

[3]

Question 12

MediumPaper 2 · calculator5 marks
(a)

Consider the function f(x)=ex−3x−6f(x) = e^x - 3x - 6.

On the following axes, sketch the graph of ff for −3≤x≤3-3 \le x \le 3.

Graph axes with x-axis from -3 to 3 and y-axis from -8 to 8, with gridlines and labels.
[3]
(b)

The function gg is defined by g(x)=e2x−6x−10g(x) = e^{2x} - 6x - 10.

The graph of gg is obtained from the graph of ff (from part a) by a horizontal stretch with scale factor kk, followed by a vertical translation of cc units.

Find the value of kk and the value of cc.

[2]

Question 13

HardPaper 1 · no calculator16 marks
(a)

A particle moves in a straight line. Its velocity, v ms−1v \,\text{ms}^{-1}, at time tt seconds is given by v(t)=−t3+6t2−9tv(t) = -t^3 + 6t^2 - 9t, for 0≤t≤50 \le t \le 5. The particle is at the origin at t=0t=0.

The graph of vv is shown in the following diagram.

Graph of velocity v against time t, showing a cubic function starting at (0,0), going down to a minimum between t=0 and t=3, then up to touch the t-axis at t=3, and then continuing downwards.

(a) Find the displacement of the particle from the origin at t=3t=3.

[4]
(b)

(b) Find an expression for the acceleration of the particle.

[2]
(c)

(c) The particle is momentarily at rest at t=0t=0 and again at t=kt=k. Find the greatest speed of the particle in the interval 0≤t≤k0 \le t \le k.

[5]
(d)

(d) Find the greatest speed of the particle for 0≤t≤50 \le t \le 5.

[2]
(e)

(e) Write down an expression that represents the distance travelled by the particle while its speed is increasing. Do not evaluate the expression.

[3]

Question 14

MediumPaper 2 · calculator14 marks
(a)

An engineer is designing an open-top rectangular container with a square base. The container must have a volume of 4 m34\text{ m}^3.

Let the side length of the square base be xx metres and the height of the container be hh metres.

Show that the total surface area, S m2S\text{ m}^2, of the material used for the container is given by S=x2+16xS = x^2 + \frac{16}{x}.

[3]
(b)(i)

Find an expression for dSdx\frac{dS}{dx}.

[2]
(b)(ii)

Hence, find the exact value of xx for which the surface area is a local minimum or maximum.

[3]
(c)(i)

Find an expression for d2Sdx2\frac{d^2S}{dx^2}.

[2]
(c)(ii)

Use the second derivative of SS to justify that SS is a minimum when x=2x = 2.

[2]
(c)(iii)

Find the minimum surface area of the container.

[2]

Question 15

HardPaper 2 · calculator20 marks
(a)

A civil engineer is analyzing the structural integrity of a new bridge design. The deflection of a certain point on the bridge, D(x)D(x), in millimeters, is modeled by the function D(x)=2x+1x2−4D(x) = \frac{2x+1}{x^2-4}, where xx represents the horizontal distance in meters from a central support. The model is valid for x∈Rx \in \mathbb{R}, x≠px\neq p, x≠qx\neq q.

Find the value of pp and the value of qq.

[2]
(b)

Find an expression for D′(x)D'(x).

[3]
(c)

The graph of y=D(x)y = D(x) has exactly one point of inflexion.

Find the x-coordinate of the point of inflexion.

[2]
(d)

Sketch the graph of y=D(x)y = D(x) for −4≤x≤4-4 \leq x \leq 4, showing the values of any axes intercepts, the coordinates of any local maxima and local minima (if they exist), and giving the equations of any asymptotes.

[5]
(e)

Consider a related model for stress distribution, S(x)=x2−42x+1S(x) = \frac{x^2-4}{2x+1} for x∈Rx \in \mathbb{R}, x≠−12x \neq -\frac{1}{2}.

Find the equations of all the asymptotes on the graph of y=S(x)y = S(x).

[4]
(f)

The engineer needs to identify the regions where the bridge deflection D(x)D(x) is less than 11 mm. Solve D(x)<1D(x) < 1 for x∈Rx \in \mathbb{R}.

[4]

Question 16

MediumPaper 2 · calculator14 marks
(a)

A packaging company is designing a new cylindrical can. The can must have a fixed volume of 54π54\pi cm3^3. The company wants to minimize the amount of material used, which corresponds to minimizing the total surface area of the can.

Let the radius of the can be rr cm and its height be hh cm.

Show that the total surface area, SS cm2^2, of the can is given by S=2πr2+108πrS = 2\pi r^2 + \frac{108\pi}{r}.

[3]
(b)(i)

The total surface area of the can has a local minimum value when r=ar = a.

(i) Find an expression for dSdr\frac{dS}{dr}.

[2]
(b)(ii)

(ii) Hence, find the exact value of aa.

[3]
(c)(i)

(i) Find an expression for d2Sdr2\frac{d^2S}{dr^2}.

[2]
(c)(ii)

(ii) Use the second derivative of SS to justify that SS is a minimum when r=ar = a.

[2]
(c)(iii)

(iii) Find the minimum surface area of the can.

[2]

Question 17

HardPaper 2 · calculator20 marks
(a)

Two drones, Drone X and Drone Y, have position vectors with respect to an origin O given respectively by

rX=(10−22)+t(−413)\boldsymbol{r}_X = \begin{pmatrix} 10 \\ -2 \\ 2 \end{pmatrix} + t \begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix}

rY=(−1−29)+t(32−1)\boldsymbol{r}_Y = \begin{pmatrix} -1 \\ -2 \\ 9 \end{pmatrix} + t \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}

where tt represents the time in minutes and 0≤t≤30 \le t \le 3.

Entries in each column vector give the displacement east of O, the displacement north of O and the distance above sea level, all measured in kilometres.

(a) Find the three-figure bearing on which Drone Y is travelling.

[2]
(b)

(b) Show that Drone X travels at a greater speed than Drone Y.

[2]
(c)

(c) Find the acute angle between the two drones' lines of flight. Give your answer in degrees.

[4]
(d)(i)

The two drones' lines of flight cross at point P.

(d) (i) Find the coordinates of P.

[5]
(d)(ii)

(ii) Determine the length of time between the first drone arriving at P and the second drone arriving at P.

[2]
(e)

(e) Let D(t)D(t) represent the distance between Drone X and Drone Y for 0≤t≤30 \le t \le 3.

Find the minimum value of D(t)D(t).

[5]

Question 18

MediumPaper 2 · calculator9 marks
(a)

A landscape architect is designing a triangular shade sail for a patio. The vertices of the sail are defined by points A, B, and C in a 3D coordinate system, where the z-axis represents height.

The coordinates of the vertices are A(0,k,1)(0, k, 1), B(2,1,0)(2, 1, 0) and C(k,0,3)(k, 0, 3), where kk is a positive constant representing a design parameter.

(a) Show that the vector product AB⃗×AC⃗\vec{AB} \times \vec{AC} is given by (2−3k−k−4k2−3k)\begin{pmatrix} 2-3k \\ -k-4 \\ k^2-3k \end{pmatrix}.

[4]
(b)

(b) The architect wants to minimize the tension in the sail, which is proportional to the magnitude of the vector product of two adjacent sides. Find the smallest possible value of ∣AB⃗×AC⃗∣|\vec{AB} \times \vec{AC}|.

[3]
(c)

(c) Calculate the smallest possible area of the shade sail.

[2]

Question 19

HardPaper 2 · calculator21 marks
(a)

The growth of a bacterial colony, BB, in a petri dish can be modelled by the logistic differential equation

dBdt=kB(1−BN)\frac{\text{d}B}{\text{d}t} = k B \left(1 - \frac{B}{N}\right)

where tt is the time measured in hours and k,Nk, N are positive constants.

The constant NN represents the maximum number of bacteria the petri dish can sustain indefinitely due to limited nutrients.

In the context of this bacterial growth model, interpret the meaning of dBdt\frac{\text{d}B}{\text{d}t}.

[1]
(b)

Show that d2Bdt2=k2B(1−BN)(1−2BN)\frac{\text{d}^2B}{\text{d}t^2} = k^2B\left(1-\frac{B}{N}\right)\left(1-\frac{2B}{N}\right).

[4]
(c)

Hence show that the bacterial colony will grow at its maximum rate when B=N2B = \frac{N}{2}. Justify your answer.

[5]
(d)

Hence determine the maximum value of dBdt\frac{\text{d}B}{\text{d}t} in terms of kk and NN.

[2]
(e)

Let B0B_0 be the initial number of bacteria.

By solving the logistic differential equation, show that its solution can be expressed in the form

kt=ln⁡(B(N−B0)B0(N−B))kt = \ln\left(\frac{B(N-B_0)}{B_0(N-B)}\right).

[7]
(f)

After 5 hours, the number of bacteria is 2B02B_0. It is known that N=3B0N = 3B_0.

Find the value of kk for this bacterial growth model.

[2]

Question 20

MediumPaper 2 · calculator17 marks
(a)

A toy rocket is launched vertically upwards from a platform. Its height, hh metres above the ground, tt seconds after launch, is modelled by the function h(t)=10+25t−4.9t2h(t) = 10 + 25t - 4.9t^2.

Find the initial height of the platform from which the rocket is launched.

[2]
(b)

Find the maximum height reached by the rocket.

[4]
(c)

Calculate the time it takes for the rocket to hit the ground.

[4]
(d)

Write down the domain of the function hh in the context of this real-life scenario.

[2]
(e)

Determine the length of time for which the height of the rocket is greater than 30 m30 \text{ m}.

[5]

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What does Second derivative, points of inflection, testing max / min points, optimisation cover in IB Maths AA?

Points of Inflection: Occur when f''(x) = 0 or undefined, and the concavity changes (verify by checking sign changes in f''(x)). Testing Max/Min Points:. f'(x) = 0 identifies critical points.

Is Second derivative, points of inflection, testing max / min points, optimisation SL or HL?

Both. SL and HL students study Second derivative, points of inflection, testing max / min points, optimisation, and HL goes further: Extended at AHL 5.12 (higher derivatives, f^(n)(x)).

How do I revise Second derivative, points of inflection, testing max / min points, optimisation for IB Maths AA?

Start from the core idea: points of Inflection: Occur when f''(x) = 0 or undefined, and the concavity changes (verify by checking sign changes in f''(x)). In the exam: "Here is the graph of f', sketch f" and its variants are the reasoning question and they are hard for students. Zeros of f' are stationary points of f, zeros of f'' with a sign change are inflexion points of f. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

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