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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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 \(f\).

M17/5/MATME/SP1/ENG/TZ1/06

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.
Answer/Explanation

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)

M17/5/MATME/SP1/ENG/TZ1/06.b/M

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.
Answer/Explanation

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.

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