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Assignment 1 solutions for calculus 1B.
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Assignment 1 solutions for calculus 1B for semester 1, 2021.
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Assignment 1.1. Let f be the function
f (x) = x2 ln x8 where x > 1
(a) Use the sign pattern for f 0 (x) to determine the intervals where f rises and where
f falls. (5)
(b) Determine the coordinates of the local extreme point(s). (2)
(c) Find f 00 (x) and determine where the graph of f is concave up and where it is
concave down. (5)
[12]
2. You are designing a poster to contain 50 cm2 of printing with margins of 4 cm each at
the top and bottom and 2 cm at each side. What overall dimensions will minimize
the amount of paper used? [5]
3. Use L’Hôpital’s rule to evaluate:
1 cos3 x
(a) lim (5)
x!0 sin2 x
(b) lim (cos x1 )x (5)
x!1
x ln x
(c) lim 2
x!1 x 1
(5)
(d) lim x2ln x (5)
x!1 x 1
[20]
1 1
4. Find the exact value of tan sin 2
[3]
TOTAL: [40]
SOLUTIONS
1. Let f be the function de…ned by
f (x) = x2 ln x8 where x > 1
Solutions:
(a)
1
, 0
f 0 (x) = (x2 8 ln x)
8
= 2x
x
2(x2 4)
=
x
2(x 2)(x + 2)
= where x > 1
x
We …nd the sign pattern for x > 1 so x in the denominator always has positive values and
may be excluded from the sign pattern. We only look at the sign pattern for x > 1
y=x 2 +
y =x+2 + +
f 0 (x) +
1 2
Thus f rises on (2; 1) and f falls on (1; 2):
(b)
The local extreme point(s) occur at x when f 0 (x) = 0:Thus we must …nd where
2(x 2)(x + 2)
f 0 (x) = = 0 in the interval x > 1
x
This can only happen when x = 2 and the local extreme point is (2; f (2)) = (2; 22 ln 28 ) =
(2; 4 8 ln 2) or (2; 4 ln 256):
(c)
00 8
f (x) = 2 + for x > 1
x2
2x2 + 8
= for x > 1 (*)
x2
We see that y = 2x2 + 8 in (*) is a parabola which have no roots and is always positive,
Similarly y = x2 is always positive for x > 1 .
Thus the second derivative is always positive and so the function is concave up on the interval
(1; 1):
2
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