IB Math Analysis & Approaches Question bank-Topic: SL 5.8 Points of inflexion with zero and non-zero gradients SL Paper 1

Question

Consider \(f(x) = \frac{1}{3}{x^3} + 2{x^2} – 5x\) . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.


Find \(f'(x)\) .

[3]
a.

Find the x-coordinate of M.

[4]
b.

Find the x-coordinate of N.

[3]
c.

The line L is the tangent to the curve of f at \((3{\text{, }}12)\). Find the equation of L in the form \(y = ax + b\) .

[4]
d.
Answer/Explanation

Markscheme

\(f'(x) = {x^2} + 4x – 5\)     A1A1A1     N3

[3 marks]

a.

evidence of attempting to solve \(f'(x) = 0\)     (M1)

evidence of correct working     A1

e.g. \((x + 5)(x – 1)\) , \(\frac{{ – 4 \pm \sqrt {16 + 20} }}{2}\) , sketch

\(x = – 5\), \(x = 1\)     (A1)

so \(x = – 5\)     A1     N2

[4 marks]

b.

METHOD 1

\(f”(x) = 2x + 4\) (may be seen later)     A1

evidence of setting second derivative = 0     (M1)

e.g. \(2x + 4 = 0\)

\(x = – 2\)     A1     N2

METHOD 2

evidence of use of symmetry     (M1)

e.g. midpoint of max/min, reference to shape of cubic

correct calculation     A1

e.g. \(\frac{{ – 5 + 1}}{2}\)

\(x = – 2\)     A1     N2

[3 marks]

c.

attempting to find the value of the derivative when \(x = 3\)     (M1)

\(f'(3) = 16\)     A1

valid approach to finding the equation of a line     M1

e.g. \(y – 12 = 16(x – 3)\) , \(12 = 16 \times 3 + b\)

\(y = 16x – 36\)     A1     N2

[4 marks]

d.

Question

The diagram shows part of the graph of \(y = f'(x)\) . The x-intercepts are at points A and C. There is a minimum at B, and a maximum at D.


(i)     Write down the value of \(f'(x)\) at C.

(ii)    Hence, show that C corresponds to a minimum on the graph of f , i.e. it has the same x-coordinate.

 

[3]
a(i) and (ii).

Which of the points A, B, D corresponds to a maximum on the graph of f ?

[1]
b.

Show that B corresponds to a point of inflexion on the graph of f .

[3]
c.
Answer/Explanation

Markscheme

(i) \(f'(x) = 0\)     A1     N1

(ii) METHOD 1

 \(f'(x) < 0\) to the left of C, \(f'(x) > 0\) to the right of C     R1R1     N2

METHOD 2

\(f”(x) > 0\)     R2     N2

[3 marks]

a(i) and (ii).

A     A1     N1

[1 mark]

b.

METHOD 1

\(f”(x) = 0\)     R2

discussion of sign change of \(f”(x)\)     R1

e.g. \(f”(x) < 0\) to the left of B and \(f”(x) > 0\) to the right of B; \(f”(x)\) changes sign either side of B

B is a point of inflexion     AG     N0

METHOD 2

B is a minimum on the graph of the derivative \({f’}\)     R2

discussion of sign change of \(f”(x)\)     R1

e.g. \(f”(x) < 0\) to the left of B and \(f”(x) > 0\) to the right of B; \(f”(x)\) changes sign either side of B

B is a point of inflexion     AG     N0

[3 marks]

c.

Question

Let \(f(x) = \frac{{ax}}{{{x^2} + 1}}\) , \( – 8 \le x \le 8\) , \(a \in \mathbb{R}\) .The graph of f is shown below.


The region between \(x = 3\) and \(x = 7\) is shaded.

Show that \(f( – x) = – f(x)\) .

[2]
a.

Given that \(f”(x) = \frac{{2ax({x^2} – 3)}}{{{{({x^2} + 1)}^3}}}\) , find the coordinates of all points of inflexion.

[7]
b.

It is given that \(\int {f(x){\rm{d}}x = \frac{a}{2}} \ln ({x^2} + 1) + C\) .

(i)     Find the area of the shaded region, giving your answer in the form \(p\ln q\) .

(ii)    Find the value of \(\int_4^8 {2f(x – 1){\rm{d}}x} \) .

[7]
c.
Answer/Explanation

Markscheme

METHOD 1

evidence of substituting \( – x\) for \(x\)     (M1)

\(f( – x) = \frac{{a( – x)}}{{{{( – x)}^2} + 1}}\)     A1

\(f( – x) = \frac{{ – ax}}{{{x^2} + 1}}\) \(( = – f(x))\)     AG     N0

METHOD 2

\(y = – f(x)\) is reflection of \(y = f(x)\) in x axis

and \(y = f( – x)\) is reflection of \(y = f(x)\) in y axis     (M1)

sketch showing these are the same     A1

\(f( – x) = \frac{{ – ax}}{{{x^2} + 1}}\) \(( = – f(x))\)     AG     N0

[2 marks]

a.

evidence of appropriate approach     (M1)

e.g. \(f”(x) = 0\)

to set the numerator equal to 0     (A1)

e.g. \(2ax({x^2} – 3) = 0\) ; \(({x^2} – 3) = 0\)

(0, 0) , \(\left( {\sqrt 3 ,\frac{{a\sqrt 3 }}{4}} \right)\) , \(\left( { – \sqrt 3 , – \frac{{a\sqrt 3 }}{4}} \right)\) (accept \(x = 0\) , \(y = 0\) etc)      A1A1A1A1A1     N5

[7 marks]

b.

(i) correct expression     A2

e.g. \(\left[ {\frac{a}{2}\ln ({x^2} + 1)} \right]_3^7\) , \(\frac{a}{2}\ln 50 – \frac{a}{2}\ln 10\) , \(\frac{a}{2}(\ln 50 – \ln 10)\)

area = \(\frac{a}{2}\ln 5\)     A1A1     N2

(ii) METHOD 1

recognizing the shift that does not change the area     (M1)

e.g. \(\int_4^8 {f(x – 1){\rm{d}}x}  = \int_3^7 {f(x){\rm{d}}x} \) , \(\frac{a}{2}\ln 5\)

recognizing that the factor of 2 doubles the area     (M1)

e.g. \(\int_4^8 {2f(x – 1){\rm{d}}x = } 2\int_4^8 {f(x – 1){\rm{d}}x} \) \(\left( { = 2\int_3^7 {f(x){\rm{d}}x} } \right)\)

\(\int_4^8 {2f(x – 1){\rm{d}}x = a\ln 5} \) (i.e. \(2 \times \) their answer to (c)(i))     A1     N3

METHOD 2

changing variable

let \(w = x – 1\) , so \(\frac{{{\rm{d}}w}}{{{\rm{d}}x}} = 1\)

\(2\int {f(w){\rm{d}}w = } \frac{{2a}}{2}\ln ({w^2} + 1) + c\)     (M1)

substituting correct limits

e.g. \(\left[ {a\ln \left[ {{{(x – 1)}^2} + 1} \right]} \right]_4^8\) , \(\left[ {a\ln ({w^2} + 1)} \right]_3^7\) , \(a\ln 50 – a\ln 10\)     (M1)

\(\int_4^8 {2f(x – 1){\rm{d}}x = a\ln 5} \)     A1     N3

[7 marks]

c.

Question

A function f has its first derivative given by \(f'(x) = {(x – 3)^3}\) .

Find the second derivative.

[2]
a.

Find \(f'(3)\) and \(f”(3)\) .

[1]
b.

The point P on the graph of f has x-coordinate \(3\). Explain why P is not a point of inflexion.

[2]
c.
Answer/Explanation

Markscheme

METHOD 1

\(f”(x) = 3{(x – 3)^2}\)     A2     N2

METHOD 2

attempt to expand \({(x – 3)^3}\)     (M1)

e.g. \(f'(x) = {x^3} – 9{x^2} + 27x – 27\)

\(f”(x) = 3{x^2} – 18x + 27\)     A1     N2

[2 marks]

a.

\(f'(3) = 0\) , \(f”(3) = 0\)     A1     N1

[1 mark]

b.

METHOD 1

\({f”}\) does not change sign at P     R1

evidence for this     R1     N0

METHOD 2

\({f’}\) changes sign at P so P is a maximum/minimum (i.e. not inflexion)     R1

evidence for this     R1     N0

METHOD 3

finding \(f(x) = \frac{1}{4}{(x – 3)^4} + c\) and sketching this function     R1

indicating minimum at \(x = 3\)     R1     N0 

[2 marks]

c.

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

Let \(f(x) = \frac{{\cos x}}{{\sin x}}\) , for \(\sin x \ne 0\) .

In the following table, \(f’\left( {\frac{\pi }{2}} \right) = p\) and \(f”\left( {\frac{\pi }{2}} \right) = q\) . The table also gives approximate values of \(f'(x)\) and \(f”(x)\) near \(x = \frac{\pi }{2}\) .


Use the quotient rule to show that \(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\) .

[5]
a.

Find \(f”(x)\) .

[3]
b.

Find the value of p and of q.

[3]
c.

Use information from the table to explain why there is a point of inflexion on the graph of f where \(x = \frac{\pi }{2}\) .

[2]
d.
Answer/Explanation

Markscheme

\(\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = – \sin x\) (seen anywhere)     (A1)(A1)

evidence of using the quotient rule     M1

correct substitution     A1

e.g. \(\frac{{\sin x( – \sin x) – \cos x(\cos x)}}{{{{\sin }^2}x}}\) , \(\frac{{ – {{\sin }^2}x – {{\cos }^2}x}}{{{{\sin }^2}x}}\)

\(f'(x) = \frac{{ – ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}\)     A1

\(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\)     AG      N0

[5 marks]

a.

METHOD 1

appropriate approach     (M1)

e.g. \(f'(x) = – {(\sin x)^{ – 2}}\)

\(f”(x) = 2({\sin ^{ – 3}}x)(\cos x)\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)     A1A1     N3

Note: Award A1 for \(2{\sin ^{ – 3}}x\) , A1 for \(\cos x\) .

METHOD 2

derivative of \({\sin ^2}x = 2\sin x\cos x\) (seen anywhere)     A1

evidence of choosing quotient rule     (M1)

e.g. \(u = – 1\) ,  \(v = {\sin ^2}x\) , \(f” = \frac{{{{\sin }^2}x \times 0 – ( – 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\)

\(f”(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)     A1     N3

[3 marks]

b.

evidence of substituting \(\frac{\pi }{2}\)     M1

e.g. \(\frac{{ – 1}}{{{{\sin }^2}\frac{\pi }{2}}}\) , \(\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}\)

\(p = – 1\) ,  \(q = 0\)    A1A1     N1N1

[3 marks]

c.

second derivative is zero, second derivative changes sign     R1R1     N2

[2 marks]

d.

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

Consider the functions \(f(x)\) , \(g(x)\) and \(h(x)\) . The following table gives some values associated with these functions.


The following diagram shows parts of the graphs of \(h\) and \(h”\) .


There is a point of inflexion on the graph of \(h\) at P, when \(x = 3\) .

Given that \(h(x) = f(x) \times g(x)\) ,

Write down the value of \(g(3)\) , of \(f'(3)\) , and of \(h”(2)\) .

[3]
a.

Explain why P is a point of inflexion.

[2]
b.

find the \(y\)-coordinate of P.

[2]
c.

find the equation of the normal to the graph of \(h\) at P.

[7]
d.
Answer/Explanation

Markscheme

\(g(3) = – 18\) , \(f'(3) = 1\) , \(h”(2) = – 6\)     A1A1A1     N3

[3 marks]

a.

\(h”(3) = 0\)     (A1)

valid reasoning     R1

eg   \({h”}\) changes sign at \(x = 3\) , change in concavity of \(h\) at \(x = 3\)

so P is a point of inflexion     AG     N0

[2 marks]

b.

writing \(h(3)\) as a product of \(f(3)\) and \(g(3)\)     A1

eg   \(f(3) \times g(3)\) , \(3 \times ( – 18)\)

\(h(3) = – 54\)     A1 N1

[2 marks]

c.

recognizing need to find derivative of \(h\)     (R1)

eg   \({h’}\) , \(h'(3)\)

attempt to use the product rule (do not accept \(h’ = f’ \times g’\) )     (M1)

eg   \(h’ = fg’ + gf’\) ,  \(h'(3) = f(3) \times g'(3) + g(3) \times f'(3)\)

correct substitution     (A1)

eg   \(h'(3) = 3( – 3) + ( – 18) \times 1\)

\(h'(3) = – 27\)    A1

attempt to find the gradient of the normal     (M1)

eg   \( – \frac{1}{m}\) , \( – \frac{1}{{27}}x\) 

attempt to substitute their coordinates and their normal gradient into the equation of a line     (M1)

eg   \( – 54 = \frac{1}{{27}}(3) + b\) , \(0 = \frac{1}{{27}}(3) + b\) , \(y + 54 = 27(x – 3)\) , \(y – 54 = \frac{1}{{27}}(x + 3)\)

correct equation in any form     A1     N4

eg   \(y + 54 = \frac{1}{{27}}(x – 3)\) , \(y = \frac{1}{{27}}x – 54\frac{1}{9}\)

[7 marks]

d.

Question

Let \(f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}\), for \(0 < x < 6\).

The graph of \(f\) has a maximum point at P.

The \(y\)-coordinate of P is \(\ln 27\).

Find the \(x\)-coordinate of P.

[3]
a.

Find \(f(x)\), expressing your answer as a single logarithm.

[8]
b.

The graph of \(f\) is transformed by a vertical stretch with scale factor \(\frac{1}{{\ln 3}}\). The image of P under this transformation has coordinates \((a,{\text{ }}b)\).

Find the value of \(a\) and of \(b\), where \(a,{\text{ }}b \in \mathbb{N}\).

[[N/A]]
c.
Answer/Explanation

Markscheme

recognizing \(f'(x) = 0\)     (M1)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6 – 2x = 0\)

\(x = 3\)    A1     N2

[3 marks]

a.

evidence of integration     (M1)

eg\(\,\,\,\,\,\)\(\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} } \)

using substitution     (A1)

eg\(\,\,\,\,\,\)\(\int {\frac{1}{u}{\text{d}}u} \) where \(u = 6x – {x^2}\)

correct integral     A1

eg\(\,\,\,\,\,\)\(\ln (u) + c,{\text{ }}\ln (6x – {x^2})\)

substituting \((3,{\text{ }}\ln 27)\) into their integrated expression (must have \(c\))     (M1)

eg\(\,\,\,\,\,\)\(\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(c = \ln 27 – \ln 9\)

EITHER

\(c = \ln 3\)    (A1)

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 3\)     A1     N4

OR

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9\)

correct use of a log law     (A1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9\)

\(f(x) = \ln \left( {3(6x – {x^2})} \right)\)    A1     N4

[8 marks]

b.

\(a = 3\)    A1     N1

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{{\ln 27}}{{\ln 3}}\)

correct use of log law     (A1)

eg\(\,\,\,\,\,\)\(\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27\)

\(b = 3\)    A1     N2

[4 marks]

c.

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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