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Date May 2018 Marks available 3 Reference code 18M.1.sl.TZ1.8
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 8 Adapted from N/A

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  f(x)

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

eg  x3+182x2+C,x3+9x2

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

eg  (1)3+9(1)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(x)=x3+9x28      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   b2a (must be clear this is for f′)

correct substitution      (A1)

eg   182×3

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.

Examiners report

[N/A]
a.
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b.
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c.

Syllabus sections

Topic 6 - Calculus » 6.4 » Indefinite integration as anti-differentiation.

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