# IB Math Analysis & Approaches Question bank-Topic: SL 5.7 Graphical behavior of functions SL Paper 1

## Question

The following diagram shows the graphs of the displacement, velocity and acceleration of a moving object as functions of time, t.

Complete the following table by noting which graph A, B or C corresponds to each function.

[4]
a.

Write down the value of t when the velocity is greatest.

[2]
b.

## Markscheme

A2A2     N4

[4 marks]

a.

$$t = 3$$     A2     N2

[2 marks]

b.

## Question

Let $$f(x) = 3 + \frac{{20}}{{{x^2} – 4}}$$ , for $$x \ne \pm 2$$ . The graph of f is given below.

The y-intercept is at the point A.

(i)     Find the coordinates of A.

(ii)    Show that $$f'(x) = 0$$ at A.

[7]
a.

The second derivative $$f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}$$ . Use this to

(i)     justify that the graph of f has a local maximum at A;

(ii)    explain why the graph of f does not have a point of inflexion.

[6]
b.

Describe the behaviour of the graph of $$f$$ for large $$|x|$$ .

[1]
c.

Write down the range of $$f$$ .

[2]
d.

## Markscheme

(i) coordinates of A are $$(0{\text{, }} – 2)$$     A1A1     N2

(ii) derivative of $${x^2} – 4 = 2x$$ (seen anywhere)     (A1)

evidence of correct approach     (M1)

e.g. quotient rule, chain rule

finding $$f'(x)$$     A2

e.g. $$f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)$$ , $$\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}$$

substituting $$x = 0$$ into $$f'(x)$$ (do not accept solving $$f'(x) = 0$$ )     M1

at A $$f'(x) = 0$$     AG     N0

[7 marks]

a.

(i) reference to $$f'(x) = 0$$ (seen anywhere)     (R1)

reference to $$f”(0)$$ is negative (seen anywhere)     R1

evidence of substituting $$x = 0$$ into $$f”(x)$$     M1

finding $$f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}$$ $$\left( { = – \frac{5}{2}} \right)$$     A1

then the graph must have a local maximum     AG

(ii) reference to $$f”(x) = 0$$ at point of inflexion     (R1)

recognizing that the second derivative is never 0     A1     N2

e.g. $$40(3{x^2} + 4) \ne 0$$ , $$3{x^2} + 4 \ne 0$$ , $${x^2} \ne – \frac{4}{3}$$ , the numerator is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

b.

correct (informal) statement, including reference to approaching $$y = 3$$     A1     N1

e.g. getting closer to the line $$y = 3$$ , horizontal asymptote at $$y = 3$$

[1 mark]

c.

correct inequalities, $$y \le – 2$$ , $$y > 3$$ , FT from (a)(i) and (c)     A1A1     N2

[2 marks]

d.

## Question

A function f is defined for $$– 4 \le x \le 3$$ . The graph of f is given below.

The graph has a local maximum when $$x = 0$$ , and local minima when $$x = – 3$$ , $$x = 2$$ .

Write down the x-intercepts of the graph of the derivative function, $$f’$$ .

[2]
a.

Write down all values of x for which $$f'(x)$$ is positive.

[2]
b.

At point D on the graph of f , the x-coordinate is $$– 0.5$$. Explain why $$f”(x) < 0$$ at D.

[2]
c.

## Markscheme

x-intercepts at $$– 3$$, 0, 2     A2     N2

[2 marks]

a.

$$– 3 < x < 0$$ , $$2 < x < 3$$     A1A1    N2

[2 marks]

b.

correct reasoning     R2

e.g. the graph of f is concave-down (accept convex), the first derivative is decreasing

therefore the second derivative is negative     AG

[2 marks]

c.

## Question

Consider $$f(x) = \ln ({x^4} + 1)$$ .

The second derivative is given by $$f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}$$ .

The equation $$f”(x) = 0$$ has only three solutions, when $$x = 0$$ , $$\pm \sqrt[4]{3}$$ $$( \pm 1.316 \ldots )$$ .

Find the value of $$f(0)$$ .

[2]
a.

Find the set of values of $$x$$ for which $$f$$ is increasing.

[5]
b.

(i)     Find $$f”(1)$$ .

(ii)     Hence, show that there is no point of inflexion on the graph of $$f$$ at $$x = 0$$ .

[5]
c.

There is a point of inflexion on the graph of $$f$$ at $$x = \sqrt[4]{3}$$ $$(x = 1.316 \ldots )$$ .

Sketch the graph of $$f$$ , for $$x \ge 0$$ .

[3]
d.

## Markscheme

substitute $$0$$ into $$f$$     (M1)

eg   $$\ln (0 + 1)$$ , $$\ln 1$$

$$f(0) = 0$$     A1 N2

[2 marks]

a.

$$f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}$$ (seen anywhere)     A1A1

Note: Award A1 for $$\frac{1}{{{x^4} + 1}}$$ and A1 for $$4{x^3}$$ .

recognizing $$f$$ increasing where $$f'(x) > 0$$ (seen anywhere)     R1

eg   $$f'(x) > 0$$ , diagram of signs

attempt to solve $$f'(x) > 0$$     (M1)

eg   $$4{x^3} = 0$$ , $${x^3} > 0$$

$$f$$ increasing for $$x > 0$$ (accept $$x \ge 0$$ )     A1     N1

[5 marks]

b.

(i)     substituting $$x = 1$$ into $$f”$$     (A1)

eg   $$\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}$$ , $$\frac{{4 \times 2}}{4}$$

$$f”(1) = 2$$     A1     N2

(ii)     valid interpretation of point of inflexion (seen anywhere)     R1

eg   no change of sign in $$f”(x)$$ , no change in concavity,

$$f’$$ increasing both sides of zero

attempt to find $$f”(x)$$ for $$x < 0$$     (M1)

eg   $$f”( – 1)$$ , $$\frac{{4{{( – 1)}^2}(3 – {{( – 1)}^4})}}{{{{({{( – 1)}^4} + 1)}^2}}}$$ , diagram of signs

correct working leading to positive value     A1

eg   $$f”( – 1) = 2$$ , discussing signs of numerator and denominator

there is no point of inflexion at $$x = 0$$     AG     N0

[5 marks]

c.

A1A1A1     N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

Only if this A1 is awarded, then award the following:

A1 for curve through ($$0$$, $$0$$) , A1 for increasing throughout.

Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

d.

## Question

The following diagram shows part of the graph of $$y = f(x)$$.

The graph has a local maximum at $$A$$, where $$x = – 2$$, and a local minimum at $$B$$, where $$x = 6$$.

On the following axes, sketch the graph of $$y = f'(x)$$.

[4]
a.

Write down the following in order from least to greatest: $$f(0),{\text{ }}f'(6),{\text{ }}f”( – 2)$$.

[2]
b.

## Markscheme

A1A1A1A1     N4

Note: Award A1 for x-intercept in circle at $$-2$$, A1 for x-intercept in circle at $$6$$.

Award A1 for approximately correct shape.

Only if this A1 is awarded, award A1 for a negative y-intercept.

[4 marks]

a.

$$f”( – 2),{\text{ }}f'(6),{\text{ }}f(0)$$     A2     N2

[2 marks]

b.

## Question

A function $$f$$ has its derivative given by $$f'(x) = 3{x^2} – 2kx – 9$$, where $$k$$ is a constant.

Find $$f”(x)$$.

[2]
a.

The graph of $$f$$ has a point of inflexion when $$x = 1$$.

Show that $$k = 3$$.

[3]
b.

Find $$f'( – 2)$$.

[2]
c.

Find the equation of the tangent to the curve of $$f$$ at $$( – 2,{\text{ }}1)$$, giving your answer in the form $$y = ax + b$$.

[4]
d.

Given that $$f'( – 1) = 0$$, explain why the graph of $$f$$ has a local maximum when $$x = – 1$$.

[3]
e.

## Markscheme

$$f”(x) = 6x – 2k$$     A1A1     N2

[2 marks]

a.

substituting $$x = 1$$ into $$f”$$     (M1)

eg$$\;\;\;f”(1),{\text{ }}6(1) – 2k$$

recognizing $$f”(x) = 0\;\;\;$$(seen anywhere)     M1

correct equation     A1

eg$$\;\;\;6 – 2k = 0$$

$$k = 3$$     AG     N0

[3 marks]

b.

correct substitution into $$f'(x)$$     (A1)

eg$$\;\;\;3{( – 2)^2} – 6( – 2) – 9$$

$$f'( – 2) = 15$$     A1     N2

[2 marks]

c.

recognizing gradient value (may be seen in equation)     M1

eg$$\;\;\;a = 15,{\text{ }}y = 15x + b$$

attempt to substitute $$( – 2,{\text{ }}1)$$ into equation of a straight line     M1

eg$$\;\;\;1 = 15( – 2) + b,{\text{ }}(y – 1) = m(x + 2),{\text{ }}(y + 2) = 15(x – 1)$$

correct working     (A1)

eg$$\;\;\;31 = b,{\text{ }}y = 15x + 30 + 1$$

$$y = 15x + 31$$     A1     N2

[4 marks]

d.

METHOD 1 ($${{\text{2}}^{{\text{nd}}}}$$ derivative)

recognizing $$f” < 0\;\;\;$$(seen anywhere)     R1

substituting $$x = – 1$$ into $$f”$$     (M1)

eg$$\;\;\;f”( – 1),{\text{ }}6( – 1) – 6$$

$$f”( – 1) = – 12$$     A1

therefore the graph of $$f$$ has a local maximum when $$x = – 1$$     AG     N0

METHOD 2 ($${{\text{1}}^{{\text{st}}}}$$ derivative)

recognizing change of sign of $$f'(x)\;\;\;$$(seen anywhere)     R1

eg$$\;\;\;$$sign chart$$\;\;\;$$

correct value of $$f’$$ for $$– 1 < x < 3$$     A1

eg$$\;\;\;f'(0) = – 9$$

correct value of $$f’$$ for $$x$$ value to the left of $$– 1$$     A1

eg$$\;\;\;f'( – 2) = 15$$

therefore the graph of $$f$$ has a local maximum when $$x = – 1$$     AG     N0

[3 marks]

Total [14 marks]

e.

## Question

Let $$y = f(x)$$, for $$– 0.5 \le$$ x $$\le$$ $$6.5$$. The following diagram shows the graph of $$f’$$, the derivative of $$f$$.

The graph of $$f’$$ has a local maximum when $$x = 2$$, a local minimum when $$x = 4$$, and it crosses the $$x$$-axis at the point $$(5,{\text{ }}0)$$.

Explain why the graph of $$f$$ has a local minimum when $$x = 5$$.

[2]
a.

Find the set of values of $$x$$ for which the graph of $$f$$ is concave down.

[2]
b.

The following diagram shows the shaded regions $$A$$, $$B$$ and $$C$$.

The regions are enclosed by the graph of $$f’$$, the $$x$$-axis, the $$y$$-axis, and the line $$x = 6$$.

The area of region $$A$$ is 12, the area of region $$B$$ is 6.75 and the area of region $$C$$ is 6.75.

Given that $$f(0) = 14$$, find $$f(6)$$.

[5]
c.

The following diagram shows the shaded regions $$A$$, $$B$$ and $$C$$.

The regions are enclosed by the graph of $$f’$$, the x-axis, the y-axis, and the line $$x = 6$$.

The area of region $$A$$ is 12, the area of region $$B$$ is 6.75 and the area of region $$C$$ is 6.75.

Let $$g(x) = {\left( {f(x)} \right)^2}$$. Given that $$f'(6) = 16$$, find the equation of the tangent to the graph of $$g$$ at the point where $$x = 6$$.

[6]
d.

## Markscheme

METHOD 1

$$f'(5) = 0$$     (A1)

valid reasoning including reference to the graph of $$f’$$     R1

eg$$\;\;\;f’$$ changes sign from negative to positive at $$x = 5$$, labelled sign chart for $$f’$$

so $$f$$ has a local minimum at $$x = 5$$     AG     N0

Note:     It must be clear that any description is referring to the graph of $$f’$$, simply giving the conditions for a minimum without relating them to $$f’$$ does not gain the R1.

METHOD 2

$$f'(5) = 0$$     A1

valid reasoning referring to second derivative     R1

eg$$\;\;\;f”(5) > 0$$

so $$f$$ has a local minimum at $$x = 5$$     AG     N0

[2 marks]

a.

attempt to find relevant interval     (M1)

eg$$\;\;\;f’$$ is decreasing, gradient of $$f’$$ is negative, $$f” < 0$$

$$2 < x < 4\;\;\;$$(accept “between 2 and 4”)     A1     N2

Notes:     If no other working shown, award M1A0 for incorrect inequalities such as $$2 \le$$ $$x$$ $$\le$$ 4, or “from 2 to 4”

[2 marks]

b.

METHOD 1 (one integral)

correct application of Fundamental Theorem of Calculus     (A1)

eg$$\;\;\;\int_0^6 {f'(x){\text{d}}x = } f(6) – f(0),{\text{ }}f(6) = 14 + \int_0^6 {f'(x){\text{d}}x}$$

attempt to link definite integral with areas     (M1)

eg$$\;\;\;\int_0^6 {f'(x){\text{d}}x = – 12 – 6.75 + 6.75,{\text{ }}\int_0^6 {f'(x){\text{d}}x = {\text{Area }}A + {\text{Area }}B + {\text{ Area }}C} }$$

correct value for $$\int_0^6 {f'(x){\text{d}}x}$$     (A1)

eg$$\;\;\;\int_0^6 {f'(x){\text{d}}x} = – 12$$

correct working     A1

eg$$\;\;\;f(6) – 14 = – 12,{\text{ }}f(6) = – 12 + f(0)$$

$$f(6) = 2$$     A1     N3

METHOD 2 (more than one integral)

correct application of Fundamental Theorem of Calculus     (A1)

eg$$\;\;\;\int_0^2 {f'(x){\text{d}}x} = f(2) – f(0),{\text{ }}f(2) = 14 + \int_0^2 {f'(x)}$$

attempt to link definite integrals with areas     (M1)

eg$$\;\;\;\int_0^2 {f'(x){\text{d}}x} = 12,{\text{ }}\int_2^5 {f'(x){\text{d}}x = – 6.75} ,{\text{ }}\int_0^6 {f'(x)} = 0$$

correct values for integrals     (A1)

eg$$\;\;\;\int_0^2 {f'(x){\text{d}}x} = – 12,{\text{ }}\int_5^2 {f'(x)} {\text{d}}x = 6.75,{\text{ }}f(6) – f(2) = 0$$

one correct intermediate value     A1

eg$$\;\;\;f(2) = 2,{\text{ }}f(5) = – 4.75$$

$$f(6) = 2$$     A1     N3

[5 marks]

c.

correct calculation of $$g(6)$$ (seen anywhere)     A1

eg$$\;\;\;{2^2},{\text{ }}g(6) = 4$$

choosing chain rule or product rule     (M1)

eg$$\;\;\;g’\left( {f(x)} \right)f'(x),{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}f(x)f'(x) + f'(x)f(x)$$

correct derivative     (A1)

eg$$\;\;\;g'(x) = 2f(x)f'(x),{\text{ }}f(x)f'(x) + f'(x)f(x)$$

correct calculation of $$g'(6)$$ (seen anywhere)     A1

eg$$\;\;\;2(2)(16),{\text{ }}g'(6) = 64$$

attempt to substitute their values of $$g'(6)$$ and $$g(6)$$ (in any order) into equation of a line     (M1)

eg$$\;\;\;{2^2} = (2 \times 2 \times 16)6 + b,{\text{ }}y – 6 = 64(x – 4)$$

correct equation in any form     A1     N2

eg$$\;\;\;y – 4 = 64(x – 6),{\text{ }}y = 64x – 380$$

[6 marks]

[Total 15 marks]

d.

## Question

The following diagram shows the graph of $$f’$$, the derivative of $$f$$.

The graph of $$f’$$ has a local minimum at A, a local maximum at B and passes through $$(4,{\text{ }} – 2)$$.

The point $${\text{P}}(4,{\text{ }}3)$$ lies on the graph of the function, $$f$$.

Write down the gradient of the curve of $$f$$ at P.

[1]
a.i.

Find the equation of the normal to the curve of $$f$$ at P.

[3]
a.ii.

Determine the concavity of the graph of $$f$$ when $$4 < x < 5$$ and justify your answer.

[2]
b.

## Markscheme

$$– 2$$     A1     N1

[1 mark]

a.i.

gradient of normal $$= \frac{1}{2}$$     (A1)

attempt to substitute their normal gradient and coordinates of P (in any order)     (M1)

eg$$\,\,\,\,\,$$$$y – 4 = \frac{1}{2}(x – 3),{\text{ }}3 = \frac{1}{2}(4) + b,{\text{ }}b = 1$$

$$y – 3 = \frac{1}{2}(x – 4),{\text{ }}y = \frac{1}{2}x + 1,{\text{ }}x – 2y + 2 = 0$$     A1     N3

[3 marks]

a.ii.

correct answer and valid reasoning     A2     N2

answer:     eg     graph of $$f$$ is concave up, concavity is positive (between $$4 < x < 5$$)

reason:     eg     slope of $$f’$$ is positive, $$f’$$ is increasing, $$f’’ > 0$$,

sign chart (must clearly be for $$f’’$$ and show A and B)

Note:     The reason given must refer to a specific function/graph. Referring to “the graph” or “it” is not sufficient.

[2 marks]

b.

## Question

A function f (x) has derivative f ′(x) = 3x2 + 18x. The graph of f has an x-intercept at x = −1.

Find f (x).

[6]
a.

The graph of f has a point of inflexion at x = p. Find p.

[4]
b.

Find the values of x for which the graph of f is concave-down.

[3]
c.

## Markscheme

evidence of integration       (M1)

eg  $$\int {f’\left( x \right)}$$

correct integration (accept absence of C)       (A1)(A1)

eg  $${x^3} + \frac{{18}}{2}{x^2} + C,\,\,{x^3} + 9{x^2}$$

attempt to substitute x = −1 into their = 0 (must have C)      M1

eg  $${\left( { – 1} \right)^3} + 9{\left( { – 1} \right)^2} + C = 0,\,\, – 1 + 9 + C = 0$$

Note: Award M0 if they substitute into original or differentiated function.

correct working       (A1)

eg  $$8 + C = 0,\,\,\,C = – 8$$

$$f\left( x \right) = {x^3} + 9{x^2} – 8$$      A1 N5

[6 marks]

a.

METHOD 1 (using 2nd derivative)

recognizing that f” = 0 (seen anywhere)      M1

correct expression for f”      (A1)

eg   6x + 18, 6p + 18

correct working      (A1)

6+ 18 = 0

p = −3       A1 N3

METHOD 1 (using 1st derivative)

recognizing the vertex of f′ is needed       (M2)

eg   $$– \frac{b}{{2a}}$$ (must be clear this is for f′)

correct substitution      (A1)

eg   $$\frac{{ – 18}}{{2 \times 3}}$$

p = −3       A1 N3

[4 marks]

b.

valid attempt to use f” (x) to determine concavity      (M1)

eg   f” (x) < 0, f” (−2), f” (−4),  6x + 18 ≤ 0

correct working       (A1)

eg   6x + 18 < 0, f” (−2) = 6, f” (−4) = −6

f concave down for x < −3 (do not accept ≤ −3)       A1 N2

[3 marks]

c.