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Topic 5.05 · SL and HL

Max / min points and solving f’(x)=0: notes and practice questions

Summary
  • Stationary points: points on a curve where the gradient is zero (f′(x)=0f'(x) = 0 or dydx=0\frac{dy}{dx} = 0).
  • Turning points: stationary points where the curve changes direction.
  • Local maximum: a turning point where the gradient changes from positive to negative.
  • Local minimum: a turning point where the gradient changes from negative to positive.
  • Point of inflection: a stationary point where the gradient is zero but the curve does not change direction.
  • Modelling: use differentiation to find maximum or minimum values of real-world quantities.
  • Steps to find and classify stationary points algebraically:
  • Differentiate the function to find f′(x)f'(x).
  • Set f′(x)=0f'(x) = 0 and solve for the x-coordinate(s).
  • Substitute x-values into the original function f(x)f(x) to find the y-coordinate(s).
  • Determine the nature (max/min) using a GDC, the first derivative test, or the second derivative test (HL).
  • Second derivative test (HL only) for determining the nature of a stationary point:
  • If f′′(x)>0f''(x) > 0, the point is a local minimum.
  • If f′′(x)<0f''(x) < 0, the point is a local maximum.
  • GDC usage:
  • Use the equation solver to find roots of f′(x)=0f'(x) = 0.
  • Use the built-in max/min feature on the graph of y=f(x)y = f(x).
  • Always use GDC to check work, unless an algebraic method is explicitly required.
  • SL/HL distinctions:
  • Finding max/min using f′(x)=0f'(x) = 0 is covered in both SL and HL.
  • The second derivative test (f′′(x)f''(x)) and formal identification of points of inflection are HL only.
  • Exam tips:
  • "Classify turning points" means to state whether they are local maximums or local minimums.
  • Substitute x-coordinates into the original function f(x)f(x) (not f′(x)f'(x)) to find y-coordinates.
  • For "show that" modelling questions, clearly show all algebraic substitution steps.

How it is examined

The local against global distinction is a stated understanding, so a question can give a restricted domain and ask for the greatest value, where the answer is at an endpoint rather than at the stationary point. At SL the nature of the point is justified from the graph or from the sign of f′(x)f'(x) either side, never from f′′(x)f''(x).

Key ideas
  • Values of xx where the gradient of a curve is zero.
  • The solution of f′(x)=0f'(x) = 0.
  • Local maximum and minimum points.

Linking questions

  • Other contexts: profit, area, volume, cost.
  • Links to other subjects: displacement-time and velocity-time graphs, and simple harmonic motion graphs (physics).
  • TOK: is it possible for an area of knowledge to describe the world without transforming it?

Practice questions

60 questions · 38 medium · 22 hard
Showing 20 of 20

Question 1

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 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)

A company is designing a closed cylindrical container to hold a specific volume of liquid. The total surface area of the container, in cm2^2, with a fixed volume of 16π16\pi cm3^3 and a radius of rr cm, is given by the function A(r)=2πr2+32πrA(r) = 2\pi r^2 + \frac{32\pi}{r}, where r>0r > 0.

Find A′(r)A'(r).

[3]
(b)(i)

Solve A′(r)=0A'(r) = 0.

[2]
(b)(ii)

Interpret your answer to (b)(i) in context.

[1]

Question 4

HardPaper 2 · calculator13 marks
(a)

(a) The position of a reconnaissance drone, relative to a control tower, is given by the vector equation r=(72)+t(−34)r = \begin{pmatrix} 7 \\ 2 \end{pmatrix} + t \begin{pmatrix} -3 \\ 4 \end{pmatrix}, where rr is the position vector in metres and tt is the time in minutes.

Write down the position vector of the drone when t=0t = 0 and when t=1t = 1.

[2]
(b)

(b) Calculate the speed of the drone.

[3]
(c)

(c) Find an expression for the distance of the drone from the origin at time tt.

[3]
(d)

(d) Hence find the minimum distance of the drone from the origin and the time at which it occurs.

[5]

Question 5

MediumPaper 1 · calculator9 marks
(a)

A landscape architect is designing a modular planter box for a new urban garden. The planter box has a base and top that are identical sectors of a circle, each with radius rr cm and angle θ\theta radians. The height of the planter box is h=2h = 2 cm. The total length of metal frame used for all edges of the planter box is L=20L = 20 cm. This includes two circular arcs, four radial edges for the top and bottom sectors, and three vertical connecting edges.

(a) Show that r=72+θr = \frac{7}{2+\theta}.

[2]
(b)(i)

(b) The planter box is designed to hold soil, enclosing a volume, VV.

(i) Find an expression for VV in terms of θ\theta.

[2]
(b)(ii)

(ii) Find the expression for dVdθ\frac{dV}{d\theta}.

[3]
(b)(iii)

(iii) Solve algebraically dVdθ=0\frac{dV}{d\theta} = 0 to find the value of θ\theta that will maximize the volume, VV.

[2]

Question 6

HardPaper 1 · calculator10 marks
(a)

The concentration of a reactant A, in mg/L, in a chemical reaction over time tt (in minutes) is modelled by the function:

C(t)=−t3+15t2−48t+100C(t) = -t^3 + 15t^2 - 48t + 100, for 0≤t≤100 \le t \le 10.

(a) Find the coordinates of the local minimum point of the concentration.

[3]
(b)

(b) Find the coordinates of the local maximum point of the concentration.

[3]
(c)

(c) Find the set of values of tt for which the concentration of reactant A is above 100100 mg/L.

[4]

Question 7

MediumPaper 1 · calculator7 marks
(a)

A company is designing a new cylindrical storage tank. The cost of manufacturing, in thousands of dollars, is modelled by the function C(r)=r2+100rC(r) = r^2 + \frac{100}{r}, where rr is the radius of the tank in metres, and r>0r > 0.

(a) Write down the equation of the vertical asymptote of C(r)C(r).

[1]
(b)

(b) Find C′(r)C'(r).

[3]
(c)

(c) Determine the interval in which C(r)C(r) is decreasing.

[3]

Question 8

HardPaper 2 · calculator22 marks
(a)

The concentration of a certain chemical, CC, in a solution over a period of time can be modelled using the function C(t)=−0.005t3+0.1t2−0.2t+5C(t) = -0.005t^3 + 0.1t^2 - 0.2t + 5, where tt is the time in hours after the experiment begins.

Sketch the graph of C(t)=−0.005t3+0.1t2−0.2t+5C(t) = -0.005t^3 + 0.1t^2 - 0.2t + 5 for 0≤t≤200 \le t \le 20.

[3]
(b)

Find the concentration after 22 hours.

[2]
(c)

Find the concentration after 1515 hours.

[2]
(d)

Find the maximum concentration and the time in hours at which this occurs.

[6]
(e)

Find the minimum concentration and the time in hours at which this occurs.

[5]
(f)

Find the times in hours when the concentration is 66 mol/L.

[4]

Question 9

MediumPaper 1 · calculator5 marks
(a)(i)

A slope field for the differential equation dydx=y−x2\frac{dy}{dx} = y - x^2 is shown.

Slope field for the differential equation dy/dx = y - x^2

Some of the solutions to the differential equation have a local maximum point and a local minimum point.

Write down the equation of the curve on which all these maximum and minimum points lie.

[2]
(a)(ii)

Sketch this curve on the slope field.

[1]
(b)

The solution to the differential equation that passes through the point (0, 1) has both a local maximum point and a local minimum point.

On the slope field, sketch the solution to the differential equation that passes through (0, 1).

[2]

Question 10

HardPaper 2 · calculator14 marks
(a)

A drone launches a package, and its trajectory is modelled by the equation h(x)=−0.015x2+0.6x+5h(x) = -0.015x^2 + 0.6x + 5, where h(x)h(x) is the height of the package in metres and xx is the horizontal distance in metres from the launch point.

On paper, sketch the graph of the path that the package flies for x≥0x \ge 0. Clearly indicate the initial height, the maximum height, and the horizontal distance when it lands.

[3]
(b)

Find the height of the package when it has travelled a horizontal distance of 1515 metres.

[2]
(c)

Find the maximum height of the package.

[4]
(d)

Find the horizontal distance at which the package lands on the ground, and explain what this value represents in the context of the problem.

[5]

Question 11

MediumPaper 1 · calculator6 marks
(a)

A company's daily production cost, C(x)C(x), in thousands of dollars, for producing xx units of a specialized component, is modelled by the function C(x)=x2+54xC(x) = x^2 + \frac{54}{x}, for x>0x > 0.

Write down the equation of the vertical asymptote of C(x)C(x).

[1]
(b)

Find C′(x)C'(x), the marginal cost function.

[3]
(c)

Determine the interval for xx where the production cost C(x)C(x) is increasing.

[2]

Question 12

HardPaper 1 · calculator9 marks
(a)

The number of visitors (in hundreds) to a new eco-tourism resort tt months after its opening is modelled by the function N(t)=−0.001t3+0.045t2−0.375t+10N(t) = -0.001 t^3 + 0.045 t^2 - 0.375 t + 10.

Sketch the graph of NN against tt for the first 4040 months, clearly indicating any intercepts and local extrema within this domain.

[2]
(b)

Find the maximum number of visitors (to the nearest whole number) during the first 4040 months.

[3]
(c)

Find the time(s) when the number of visitors is above 12001200. Give your answer in months, correct to two decimal places.

[4]

Question 13

MediumPaper 1 · calculator6 marks
(a)

The relationship between the sound intensity, S, of a speaker and the distance, d, from the speaker can be modelled by S=kd2S = \frac{k}{d^2}.

A sound engineer measures the sound intensity at different distances. The data collected is shown in the table.

d (m)S (W/m2^2)
198
226
54.5
d (m)S (W/m2^2)
198
226
54.5

The engineer finds the sum of square residuals in the form 1.0641k2−209.36k+c1.0641k^2 - 209.36k + c.

Find the exact value of c.

[4]
(b)

Hence find the least squares regression curve of the form S=kd2S = \frac{k}{d^2}.

[2]

Question 14

HardPaper 2 · calculator13 marks
(a)

A company is designing an open-top storage container with a square base. The side length of the base is xx cm and the height is hh cm. The container needs to have a volume of 500500 cm3^3.

Explain why x2h=500x^2 h = 500.

[1]
(b)

Rearrange the equation in part (a) to make hh the subject.

[1]
(c)

Write down an expression for the surface area, AA, of the open-top container.

[2]
(d)

Show that this can be written as

A=x2+2000xA = x^2 + \frac{2000}{x}

[3]
(e)

Plot the graph of A=x2+2000xA = x^2 + \frac{2000}{x} for x>0x > 0.

[2]
(f)

Find the minimum surface area and the value of xx when this occurs.

[4]

Question 15

MediumPaper 1 · calculator9 marks
(a)

(a) A drone's vertical velocity, vv metres per second, at time tt seconds, is given by v=tsin⁡(2t2)v = t \sin(2t^2).

Find an expression for the vertical acceleration of the drone.

[2]
(b)

(b) Hence, or otherwise, find its greatest vertical acceleration for 0≤t≤30 \le t \le 3 seconds.

[2]
(c)

(c) The drone starts at ground level (displacement is 0). Find an expression for the vertical displacement of the drone.

[3]
(d)

(d) Hence show that the drone never descends below ground level.

[2]

Question 16

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 17

MediumPaper 1 · calculator10 marks
(a)

A landscape architect is designing a series of modular planter boxes for an urban garden project. The volume, VV cm3^{3}, of a particular planter box is modelled by the function V(d)=30d2−d3V(d) = 30d^2 - d^3, where dd is the depth of the planter in cm.

(a) Use your graphic display calculator to find the value of dd that will produce the maximum volume.

[2]
(b)

(b) Show that the maximum volume of the planter box is 40004000 cm3^{3}.

[4]
(c)

The width of the planter is dd cm, and the length is (30−d)(30 - d) cm.

(c) The architect is interested in the total linear dimension LL, which is defined as the sum of the depth, width, and length of the planter. Hence find the value of LL when the volume is maximized.

[4]

Question 18

HardPaper 1 · calculator7 marks
(a)

(a) When the profit is zero, find the possible number of units produced, xx.

[3]
(b)

(b) Determine the positive values of profit, PP, for which there is only one positive value of xx (units produced).

[4]

Question 19

MediumPaper 1 · calculator7 marks

A sculptor is designing a curved art installation whose height above the ground, in meters, can be modelled by the function h(x)=x3−3x2+2h(x) = x^3 - 3x^2 + 2, where xx is the horizontal distance in meters from a reference point. The installation is planned for a section from x=−1x = -1 to x=3x = 3 meters.

(a) Draw the graph of the function h(x)=x3−3x2+2h(x) = x^3 - 3x^2 + 2, for −1≤x≤3-1 \le x \le 3, clearly indicating any local maximum or minimum points and the endpoints of the curve. Hence, determine the range of the function for this domain.

Question 20

HardPaper 2 · calculator15 marks
(a)

A drone is launched vertically upwards from a platform. Its vertical velocity, v ms−1v \text{ ms}^{-1}, at time tt seconds, is given by the function:

v=−3t2+18t−15v = -3t^2 + 18t - 15, for t≥0t \ge 0.

Find the times when the drone is momentarily at rest.

[2]
(b)

Find the magnitude of the drone's vertical acceleration at t=5t = 5 seconds.

[4]
(c)

Find the greatest speed of the drone in the interval 0≤t≤50 \le t \le 5.

[2]
(d)

The drone starts from an initial height of 1010 metres above the ground. Find an expression for the height of the drone, hh metres, above the ground at time tt seconds.

[4]
(e)

Find the total distance travelled by the drone in the interval 0≤t≤40 \le t \le 4.

[3]

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What does Max / min points and solving f’(x)=0 cover in IB Maths AI?

Stationary points: points on a curve where the gradient is zero (f'(x) = 0 or (dy)/(dx) = 0). Turning points: stationary points where the curve changes direction. Local maximum: a turning point where the gradient changes from positive to negative.

Is Max / min points and solving f’(x)=0 SL or HL?

Both. SL and HL students study Max / min points and solving f’(x)=0 to the same depth.

How do I revise Max / min points and solving f’(x)=0 for IB Maths AI?

Start from the core idea: stationary points: points on a curve where the gradient is zero (f'(x) = 0 or (dy)/(dx) = 0). In the exam: the local against global distinction is a stated understanding, so a question can give a restricted domain and ask for the greatest value, where the answer is at an endpoint rather than at the stationary point. At SL the nature of the point is justified from the graph or from the sign of f'(x) either side, never from f''(x). Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Max / min points and solving f’(x)=0?

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Yes. In the FourtyFive iPad app you write your working by hand with Apple Pencil, the way you would on paper, and it is marked the same way.

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