MA1S11 Tutorial Sheet 9, solutions1
8-11 December 2015
Questions
1. (2) Find the absolute maximum of
1
f (x) = − x4 + x3 + 2x2 + 1
4
(1)
Solution: The function is polynomial and thus twice differentiable everywhere. We
start by computing the first and second derivative:
f ′ (x) = −x3 + 3x2 + 4x = −x(x − 4)(x + 1),
f ′′ (x) = −3x2 + 6x + 4.
(2)
The factorized form immediately gives the stationary points at x = 0, 4, −1. Evaluating the second derivatives at these points we get
f ′′ (0) = 4,
f ′′ (−1) = −5,
f ′′ (4) = −20.
(3)
from which we conclude that at x = −1 and x = 4 are relative maxima and at x = 0
a relative minimum. Evaluating the function itself at the stationary points gives
f (0) = 1,
f (−1) = 7/4,
f (4) = 33.
(4)
together with the fact that
lim f (x) = −∞
x→±∞
(5)
we see that the absolute maximum must be attained at x = 4. The graph of f (x) is
given in figure 1.
√
2. (2) Use Newton’s method to find 5 3 to within 0.01 by solving x5 = 3, with x1 = 2
as a starting value.
Solution: We define f (x) = x5 − 3, with f ′ (x) = 5x4 . This gives
xn+1 = xn −
4x5n + 3
x5n − 3
=
.
5x4n
5x4n
(6)
Starting with x1 = 2 we get successive approximations
x2
x3
x4
x5
x6
1
=
≈
≈
≈
≈
1.6375
1.393449938
1.273902301
1.246949665
1.245733320
Stefan Sint, sint@maths.tcd.ie, see also http://www.maths.tcd.ie/~sint/MA1S11.html
1
(7)
(8)
(9)
(10)
(11)
50
40
30
y
20
10
0
−5.0
−2.5
0.0
2.5
5.0
7.5
10.0
x
Figure 1: graph of f (x) in question 1.
and, since the third digit is no longer changing, we have probably reached the desired
accuracy. It is easy to check using a calculator that this does indeed give the fifth
root of three to this precision.
√
√
√
3. (4) Find the indefinite integral of x3 − 4x, x + 1/ x, x3 − 3 x and x(1 + x4 ).
Solution: All these functions can be integrated using the power rule and linearity of
the indefinite integral
Z
1 4
(x3 − 4x)dx =
x − 2x2 + C
(12)
4
Z
√
√
2 3/2
x + 2x1/2 + C
(13)
( x + 1/ x)dx =
3
Z
√
1 4 3 4/3
x − x +C
(14)
(x3 − 3 x)dx =
4
4
Z
1 2 1 6
x(1 + x4 )dx =
x + x +C
(15)
2
6
2
Extra Questions
The questions are extra; you don’t need to do them in the tutorial class.
(a) Evaluate
Z
1
dx
(16)
1 − sin x
Hint: first multiply by 1 + sin x above and below the line, then use the identity
cos2 x + sin2 x = 1.
Solution:
We first transform the integrand:
1
1 + sin x
1
1 + sin x
= sec2 x + tan x sec x.
=
=
2
1 − sin x
1 − sin x 1 + sin x
1 − sin x
From the integral table we then find
Z
Z
1
dx = (sec2 x + tan x sec x)dx = tan x + sec x + C
1 − sin x
(17)
(18)
4. Find the points of inflection of
f (x) = x4 − 6x2 + 12x + 24
(19)
Solution:
We compute the second derivative
f ′ (x) = 4x3 − 12x + 12
f ′′ (x) = 12x2 − 12 = 12(x + 1)(x − 1)
(20)
(21)
so f ′′ (x) = 0 is solved for x = 1 and x = −1. Both are inflection points as the sign
of f ′′ (x) is changing across x = −1 and across x = 1, so both points separate regions
of concavity up from regions of concavity down.
3