IB Mathematics: Analysis and Approaches Quiz: Differentiation Applications

20 questions · exam conditions

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Differentiation ApplicationsQuestion 1 of 20

The function $f$ is defined by $f(x) = x^3 - 6x^2 + 5$ . Find the set of values of $x$ for which $f(x)$ is a decreasing function.

x < 0

x > 4

0 < x < 4

x < 2

x > 2

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Differentiation Applications

Practice Differentiation Applications in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Differentiation Applications, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

The function $f$ is defined by $f(x) = x^3 - 6x^2 + 5$ . Find the set of values of $x$ for which $f(x)$ is a decreasing function.

$x < 0$ or $x > 4$
$0 < x < 4$ (correct answer)
$x < 2$
$x > 2$

Explanation: A function is decreasing when its first derivative is negative. First, find the derivative of

f(x)

f'(x) = 3x^2 - 12x

. We need to find where

f'(x) < 0

. We solve

3x^2 - 12x < 0

, which is

3x(x - 4) < 0

. The roots of the quadratic

3x(x-4)=0

are

x=0

and

x=4

. Since the quadratic is a parabola opening upwards, it will be negative between its roots. Therefore, the function is decreasing for

0 < x < 4

Question 2

The position of a particle moving along a straight line is given by $s(t) = t^3 - 12t^2 + 36t$ for $t \ge 0$ , where $t$ is in seconds and $s$ is in meters. At what time $t$ does the particle first momentarily stop?

0 s
6 s
4 s
2 s (correct answer)

Explanation: When you encounter a particle motion problem asking when the particle "momentarily stops," you need to find when the velocity equals zero. The velocity is the first derivative of the position function. Given

s(t) = t^3 - 12t^2 + 36t

, let's find the velocity by taking the derivative:

v(t) = s'(t) = 3t^2 - 24t + 36

To find when the particle stops, set the velocity equal to zero:

3t^2 - 24t + 36 = 0

Divide everything by 3 to simplify:

t^2 - 8t + 12 = 0

Factor this quadratic equation:

(t - 6)(t - 2) = 0

This gives us

t = 2

t = 6

. Since the question asks for the first time the particle stops, the answer is

t = 2

seconds, which is choice D. Let's examine the wrong answers: Choice A (0 s) occurs because at

t = 0

, the velocity is

v(0) = 36 \neq 0

, so the particle is moving initially. Choice B (6 s) is when the particle stops for the second time, not the first. Choice C (4 s) might tempt you if you made an error in factoring or solving the quadratic—at

t = 4

, we get

v(4) = -12 \neq 0

. Remember: "momentarily stops" means velocity equals zero, not position equals zero. Always take the derivative of position to get velocity, then solve

v(t) = 0

. Watch for multiple solutions and choose the first positive time when asked.

Question 3

A cylindrical can with a closed top and bottom is to be made to hold a volume of $1000 \text{ cm}^3$ . The cost of the material for the top and bottom is twice the cost per unit area of the material for the side. Find the radius $r$ that minimizes the cost of the can.

$\sqrt[3]{\frac{250}{\pi}}$ (correct answer)
$\sqrt[3]{\frac{500}{\pi}}$
$\sqrt[3]{\frac{1000}{\pi}}$
$\sqrt[3]{\frac{2000}{\pi}}$

Explanation: Let

r

be the radius and

h

be the height. The volume is

V = \pi r^2 h = 1000

, so

h = \frac{1000}{\pi r^2}

. Let the cost of the side material be

k

per cm². The cost of the top/bottom material is (2k). The area of the side is

A_s = 2\pi rh

and the area of the top and bottom is

A_{tb} = 2\pi r^2

. The total cost is

C = k A_s + 2k A_{tb} = k(2\pi rh) + 2k(2\pi r^2) = 2k\pi(rh + 2r^2)

. Substitute

h

C(r) = 2k\pi(r \frac{1000}{\pi r^2} + 2r^2) = 2k\pi(\frac{1000}{\pi r} + 2r^2) = \frac{2000k}{r} + 4k\pi r^2

. To minimize cost, find the derivative and set to zero:

C'(r) = -\frac{2000k}{r^2} + 8k\pi r

. Setting

C'(r)=0

gives

8k\pi r = \frac{2000k}{r^2}

. The

k

cancels out.

8\pi r^3 = 2000 \implies r^3 = \frac{2000}{8\pi} = \frac{250}{\pi}

. Thus,

r = \sqrt[3]{\frac{250}{\pi}}

Question 4

A rectangular field is to be enclosed by a fence. One side of the field is along a straight river and does not need fencing. The total length of the fence used for the other three sides is 200 metres. Find the maximum possible area of the field.

2500 m²
5000 m² (correct answer)
7500 m²
10000 m²

Explanation: Let the side parallel to the river have length

y

and the other two sides have length

x

. The total length of the fence is

2x + y = 200

, so

y = 200 - 2x

. The area of the field is

A = xy = x(200 - 2x) = 200x - 2x^2

. To maximize the area, we find the derivative with respect to

x

and set it to zero:

A'(x) = 200 - 4x

. Setting

A'(x) = 0

gives

200 - 4x = 0

, which solves to

x = 50

. The corresponding length of the other side is

y = 200 - 2(50) = 100

. The maximum area is

A = 50 \times 100 = 5000

m². The second derivative

A''(x) = -4 < 0

, confirming it is a maximum.

Question 5

Find the maximum value of the function $f(x) = 2x^3 - 9x^2 + 12x$ on the interval $[0, 3]$ .

0
4
5
9 (correct answer)

Explanation: To find the absolute maximum value on a closed interval, we must evaluate the function at its critical points within the interval and at the endpoints of the interval. First, find the critical points by setting the derivative to zero:

f'(x) = 6x^2 - 18x + 12

. Set

f'(x) = 0

6(x^2 - 3x + 2) = 0 \implies 6(x-1)(x-2) = 0

. The critical points are

x=1

and

x=2

, both of which are in the interval

[0, 3]

. Now, evaluate the function at the critical points and the endpoints:

f(0) = 0

f(1) = 2 - 9 + 12 = 5

f(2) = 2(8) - 9(4) + 12(2) = 16 - 36 + 24 = 4

f(3) = 2(27) - 9(9) + 12(3) = 54 - 81 + 36 = 9

Comparing the values {0, 5, 4, 9}, the maximum value is 9.

Question 6

The cost $C$ of producing $x$ items is given by $C(x) = 0.1x^3 - 6x^2 + 150x + 200$ , for $x > 0$ . Find the number of items that should be produced to minimize the marginal cost.

10
20 (correct answer)
30
40

Explanation: The marginal cost,

M(x)

, is the derivative of the cost function,

C(x)

. So,

M(x) = C'(x) = 0.3x^2 - 12x + 150

. To minimize the marginal cost, we need to find the derivative of

M(x)

and set it to zero.

M'(x) = 0.6x - 12

. Setting

M'(x) = 0

gives

0.6x - 12 = 0

, which solves to

x = 12 / 0.6 = 20

. To confirm this is a minimum, we check the second derivative:

M''(x) = 0.6 > 0

, which confirms a minimum. So, 20 items should be produced to minimize the marginal cost.

Question 7

The second derivative of a function is given by $f''(x) = (x-1)(x-3)^2$ . How many points of inflection does the graph of $y=f(x)$ have?

0
1 (correct answer)
2
3

Explanation: A point of inflection occurs where the second derivative

f''(x)

is zero and changes sign. The possible points of inflection are where

f''(x) = 0

, which occurs at

x=1

and

x=3

. We must test the sign of

f''(x)

around these points. The term

(x-3)^2

is always non-negative. Therefore, the sign of

f''(x)

is determined by the sign of the term

(x-1)

. For

x < 1

x-1

is negative, so

f''(x) < 0

. For

x > 1

(but

x \neq 3

x-1

is positive, so

f''(x) > 0

. Since the sign of

f''(x)

changes at

x=1

, it is a point of inflection. Around

x=3

, for example at

x=2.9

and

x=3.1

, the term

x-1

is positive, and

(x-3)^2

is positive. So the sign of

f''(x)

does not change at

x=3

. Thus,

x=3

is not a point of inflection. There is only one point of inflection.

Question 8

2500 m²
5000 m² (correct answer)
7500 m²
10000 m²

Explanation: Let the side parallel to the river have length

y

and the other two sides have length

x

. The total length of the fence is

2x + y = 200

, so

y = 200 - 2x

. The area of the field is

A = xy = x(200 - 2x) = 200x - 2x^2

. To maximize the area, we find the derivative with respect to

x

and set it to zero:

A'(x) = 200 - 4x

. Setting

A'(x) = 0

gives

200 - 4x = 0

, which solves to

x = 50

. The corresponding length of the other side is

y = 200 - 2(50) = 100

. The maximum area is

A = 50 \times 100 = 5000

m². The second derivative

A''(x) = -4 < 0

, confirming it is a maximum.

Question 9

A company's profit, $P$ , in thousands of dollars, from selling $x$ hundred items is given by $P(x) = -x^3 + 9x^2 - 15x$ . For which number of items sold is the profit maximized?

100
300
500 (correct answer)
700

Explanation: To find the maximum profit, we need to find the stationary points of the profit function

P(x)

by setting its derivative to zero.

P'(x) = -3x^2 + 18x - 15

. Set

P'(x) = 0

-3(x^2 - 6x + 5) = 0 \implies -3(x-1)(x-5) = 0

. The stationary points are at

x=1

and

x=5

. To determine which is a maximum, we use the second derivative test:

P''(x) = -6x + 18

. At

x=1

P''(1) = -6(1) + 18 = 12 > 0

, which indicates a local minimum. At

x=5

P''(5) = -6(5) + 18 = -30 + 18 = -12 < 0

, which indicates a local maximum. Since

x

represents hundreds of items,

x=5

corresponds to 500 items.

Question 10

$\sqrt[3]{\frac{250}{\pi}}$ (correct answer)
$\sqrt[3]{\frac{500}{\pi}}$
$\sqrt[3]{\frac{1000}{\pi}}$
$\sqrt[3]{\frac{2000}{\pi}}$

Explanation: Let

r

be the radius and

h

be the height. The volume is

V = \pi r^2 h = 1000

, so

h = \frac{1000}{\pi r^2}

. Let the cost of the side material be

k

per cm². The cost of the top/bottom material is (2k). The area of the side is

A_s = 2\pi rh

and the area of the top and bottom is

A_{tb} = 2\pi r^2

. The total cost is

C = k A_s + 2k A_{tb} = k(2\pi rh) + 2k(2\pi r^2) = 2k\pi(rh + 2r^2)

. Substitute

h

C(r) = 2k\pi(r \frac{1000}{\pi r^2} + 2r^2) = 2k\pi(\frac{1000}{\pi r} + 2r^2) = \frac{2000k}{r} + 4k\pi r^2

. To minimize cost, find the derivative and set to zero:

C'(r) = -\frac{2000k}{r^2} + 8k\pi r

. Setting

C'(r)=0

gives

8k\pi r = \frac{2000k}{r^2}

. The

k

cancels out.

8\pi r^3 = 2000 \implies r^3 = \frac{2000}{8\pi} = \frac{250}{\pi}

. Thus,

r = \sqrt[3]{\frac{250}{\pi}}

Question 11

0 s
6 s
4 s
2 s (correct answer)

s(t) = t^3 - 12t^2 + 36t

, let's find the velocity by taking the derivative:

v(t) = s'(t) = 3t^2 - 24t + 36

To find when the particle stops, set the velocity equal to zero:

3t^2 - 24t + 36 = 0

Divide everything by 3 to simplify:

t^2 - 8t + 12 = 0

Factor this quadratic equation:

(t - 6)(t - 2) = 0

This gives us

t = 2

t = 6

. Since the question asks for the first time the particle stops, the answer is

t = 2

seconds, which is choice D. Let's examine the wrong answers: Choice A (0 s) occurs because at

t = 0

, the velocity is

v(0) = 36 \neq 0

t = 4

, we get

v(4) = -12 \neq 0

. Remember: "momentarily stops" means velocity equals zero, not position equals zero. Always take the derivative of position to get velocity, then solve

v(t) = 0

. Watch for multiple solutions and choose the first positive time when asked.

Question 12

A rectangle has a perimeter of 24 cm. What is the maximum possible area of this rectangle?

24 cm²
32 cm²
35 cm²
36 cm² (correct answer)

Explanation: Let the length and width of the rectangle be

l

and

w

respectively. The perimeter is

2l + 2w = 24

, which simplifies to

l + w = 12

. We can express

w

w = 12 - l

. The area is

A = l \times w = l(12 - l) = 12l - l^2

. To find the maximum area, we take the derivative of

A

with respect to

l

and set it to zero:

A'(l) = 12 - 2l

. Setting

A'(l) = 0

gives

12 - 2l = 0

, so

l = 6

. When

l = 6

w = 12 - 6 = 6

. The rectangle with maximum area is a square. The maximum area is

A = 6 \times 6 = 36

cm².

Question 13

Find the equation of the normal to the curve $y = \ln(x^2+1)$ at the point where $x=1$ .

$y = -x + \ln(2) + 1$ (correct answer)
$y = x + \ln(2) - 1$
$y = -x + \ln(2) - 1$
$y = x + \ln(2) + 1$

Explanation: First, find the point of interest. When

x=1

y = \ln(1^2+1) = \ln(2)

. The point is

(1, \ln(2))

. Next, find the gradient of the tangent by differentiating

y

. Using the chain rule,

\frac{dy}{dx} = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}

. At

x=1

, the gradient of the tangent is

m_T = \frac{2(1)}{1^2+1} = 1

. The gradient of the normal is the negative reciprocal of the tangent's gradient, so

m_N = -1/1 = -1

. The equation of the normal is given by

y - y_1 = m_N(x - x_1)

. Substituting the point and gradient gives

y - \ln(2) = -1(x - 1)

, which simplifies to

y - \ln(2) = -x + 1

, or

y = -x + 1 + \ln(2)

Question 14

The graph of a function $y=f(x)$ is concave down for all $x<3$ and concave up for all $x>3$ . Which of the following is a possible expression for $f''(x)$ ?

$f''(x) = 3-x$
$f''(x) = (x-3)^2$
$f''(x) = x-3$ (correct answer)
$f''(x) = 9-x^2$

Explanation: The concavity of a function is determined by the sign of its second derivative. Concave down corresponds to

f''(x) < 0

, and concave up corresponds to

f''(x) > 0

. We are given that the graph is concave down for

x<3

(so

f''(x)<0

for

x<3

) and concave up for

x>3

(so

f''(x)>0

for

x>3

). We need to check which of the given expressions for

f''(x)

satisfies these conditions. A.

f''(x) = 3-x

: If

x<3

3-x>0

(concave up). Incorrect. B.

f''(x) = (x-3)^2

: This expression is non-negative for all

x

, so it doesn't describe a change from concave down to up. Incorrect. C.

f''(x) = x-3

: If

x<3

x-3<0

(concave down). If

x>3

x-3>0

(concave up). This matches the description. Correct. D.

f''(x) = 9-x^2

: This is a downward-opening parabola with roots at

x=\pm3

. It is negative for

x>3

and for

x<-3

. Incorrect.

Question 15

A company's profit, $P$ , in thousands of dollars, from selling $x$ hundred items is given by $P(x) = -x^3 + 9x^2 - 15x$ . For which number of items sold is the profit maximized?

100
300
500 (correct answer)
700

Explanation: To find the maximum profit, we need to find the stationary points of the profit function

P(x)

by setting its derivative to zero.

P'(x) = -3x^2 + 18x - 15

. Set

P'(x) = 0

-3(x^2 - 6x + 5) = 0 \implies -3(x-1)(x-5) = 0

. The stationary points are at

x=1

and

x=5

. To determine which is a maximum, we use the second derivative test:

P''(x) = -6x + 18

. At

x=1

P''(1) = -6(1) + 18 = 12 > 0

, which indicates a local minimum. At

x=5

P''(5) = -6(5) + 18 = -30 + 18 = -12 < 0

, which indicates a local maximum. Since

x

represents hundreds of items,

x=5

corresponds to 500 items.

Question 16

A function $f(x)$ has derivatives $f'(x)$ and $f''(x)$ . It is known that $f'(3) = 0$ and $f''(3) = -4$ . Which statement must be true about the graph of $y=f(x)$ at $x=3$ ?

It has a local minimum.
It has a local maximum. (correct answer)
It has a non-stationary point of inflection.
It is increasing and concave down.

Explanation: The conditions given are for the second derivative test for local extrema. The first condition,

f'(3) = 0

, indicates that there is a stationary point at

x=3

. The second condition,

f''(3) = -4 < 0

, indicates that the curve is concave down at this stationary point. A stationary point on a concave down section of a curve is a local maximum.

Question 17

The derivative of a function $f$ is given by $f'(x) = x(x-2)^2$ . At which value of $x$ does $f(x)$ have a local minimum?

$x=0$ (correct answer)
$x=1$
$x=2$
$f(x)$ has no local minimum.

Explanation: Local extrema occur at stationary points, where

f'(x) = 0

. Setting

x(x-2)^2 = 0

gives stationary points at

x=0

and

x=2

. To determine the nature of these points, we use the first derivative test. We check the sign of

f'(x)

in the intervals around the stationary points. For

x < 0

(e.g.,

x=-1

f'(-1) = (-1)(-3)^2 = -9 < 0

. For

0 < x < 2

(e.g.,

x=1

f'(1) = (1)(-1)^2 = 1 > 0

. Since the sign of

f'(x)

changes from negative to positive at

x=0

, there is a local minimum at

x=0

. For

x > 2

(e.g.,

x=3

f'(3) = (3)(1)^2 = 3 > 0

. The sign does not change at

x=2

, so it is a stationary point of inflection.

Question 18

Find the x-coordinates of the stationary points of the function $f(x) = x^4 - 4x^3 + 10$ .

$x=0$ only
$x=3$ only
$x=0$ and $x=3$ (correct answer)
$x=0$ and $x=4$

Explanation: Stationary points occur where the first derivative of the function is equal to zero. First, find the derivative of

f(x)

f'(x) = 4x^3 - 12x^2

. Now, set the derivative to zero and solve for

x

4x^3 - 12x^2 = 0

. We can factor out

4x^2

4x^2(x - 3) = 0

. The solutions are found by setting each factor to zero:

4x^2 = 0 \implies x = 0

and

x - 3 = 0 \implies x = 3

. Thus, the x-coordinates of the stationary points are 0 and 3.

Question 19

The function $f(x)$ is twice differentiable. If $f'(x) > 0$ and $f''(x) < 0$ for all $x$ in the interval $(a, b)$ , which statement best describes the graph of $f(x)$ on this interval?

Increasing and concave up
Increasing and concave down (correct answer)
Decreasing and concave up
Decreasing and concave down

Explanation: The sign of the first derivative determines whether the function is increasing or decreasing. Since

f'(x) > 0

, the function

f(x)

is increasing. The sign of the second derivative determines the concavity of the function. Since

f''(x) < 0

, the function

f(x)

is concave down. Therefore, the graph of

f(x)

is increasing and concave down on the interval

(a, b)

Question 20

A function $f(x)$ has derivatives $f'(x)$ and $f''(x)$ . It is known that $f'(3) = 0$ and $f''(3) = -4$ . Which statement must be true about the graph of $y=f(x)$ at $x=3$ ?

It has a local minimum.
It has a local maximum. (correct answer)
It has a non-stationary point of inflection.
It is increasing and concave down.

Explanation: The conditions given are for the second derivative test for local extrema. The first condition,

f'(3) = 0

, indicates that there is a stationary point at

x=3

. The second condition,

f''(3) = -4 < 0

, indicates that the curve is concave down at this stationary point. A stationary point on a concave down section of a curve is a local maximum.