Mathematics of Engineering II
SS 2016
Prof. Dr. R. Reemtsen
M. Sc. M. Bähr
Exercise Sheet 4 - Solution
Problem 4.1
inverse function f −1 (x) exists, if f (x) strictly monotonically decreasing or increasing
a) strictly monotonically increasing, this means f (x) < f (y) if x < y, because
f (x) − f (y) = (4x − 2) − (4y − 2) = 4(x − y) < 0
• inverse:
y = 4x − 2
⇐⇒
y+2
=x
4
=⇒
f −1 (x) =
b) strictly monotonically increasing, this means f (x) < f (y) if
f (x) < f (y)
⇐⇒
⇐⇒
√
4x − 3 <
x<y
3
4
x+2
4
≤ x < y, because
p
4y − 3
• inverse:
y=
√
4x − 3
⇐⇒
y2 + 3
=x
4
=⇒
Problem 4.2
a) p(x) = x5 − 21x3 + 20x2 = x2 (x3 − 21x + 20)
• first zero is x1 = 0 with multiplicity k1 = 2
• find zero of x3 − 21x + 20, for example x2 = 1
• polynomial division leads to x2 + x − 20
• zero’s are x3 = 4 and x4 = −5
=⇒ p(x) = x5 − 21x3 + 20x2 = x2 (x − 1)(x − 4)(x + 5)
b) p(x) = x3 + 2x2 + x + 2
f −1 (x) =
x2 + 3
4
• find zero, for example x1 = −2
• polynomial division leads to x2 + 1
• zero’s are x2 = i and x3 = −i
=⇒ p(x) = x3 + 2x2 + x + 2 = (x + 2)(x − i)(x + i) = (x + 2)(x2 + 1)
Problem 4.3
• r(x) =
sion
x3
,
x2 −3x+2
• this leads to
degree of numerator ≥ degree of denominator =⇒ polynomial divix+3+
7x−6
x2 −3x+2
• zero’s are x1 = 1 and x2 = 2
• approach:
7x − 6
A
B
A(x − 2) + B(x − 1)
=
+
=
(x − 1)(x − 2)
(x − 1) (x − 2)
(x − 1)(x − 2)
• 7x − 6 = A(x − 2) + B(x − 1) =⇒ A = −1 and B = 8
• therefore r(x) =
Problem 4.4
a) 3
b) 5
c) exp(x)
d) 1 + ln(x)
x3
x2 −3x+2
=x+3−
1
(x−1)
+
8
(x−2)
Homework
(due April 13/18, 2016)
Problem H 4.5
a) strictly monotonically decreasing
• inverse:
f −1 (x) =
1
x
b) strictly monotonically increasing, because for 0 < x < y
f (x) − f (y) = · · · =
2(x2 y − xy 2 ) + 2(x − y)
<0
4xy
• inverse:
y=
x2 − 1
2x
⇐⇒
=⇒
• note: x = y −
x2 − 2xy − 1 = 0
|
{z
}
find zero’s
√
f −1 (x) = x + x2 + 1
p
y 2 + 1 does not exist
Problem H 4.6
a) p(x) = x3 + 10x2 + 8x − 64 = (x − 2)(x + 4)(x + 8)
b) p(x) = x3 − 2x2 − 23x + 150
• zero’s are x1 = −6, x2 = 4 + 3i and x3 = 4 − 3i
=⇒ p(x) = x3 − 2x2 − 23x + 150 = (x + 6)(x − (4 + 3i))(x − (4 − 3i)) = (x + 6)[(x − 4)2 + 9]
Problem H 4.7
a) r(x) =
• we get
2x−4
x2 −2x−8
=
2x−4
(x−4)(x+2)
2x − 4 = A(x + 2) + B(x − 4) =⇒ A =
2
3
and B =
4
3
• therefore r(x) =
b) r(x) =
x2
x2 −2x+1
• approach:
2x−4
x2 −2x−8
=1+
2x−1
(x−1)2
=
=
2
3
x−4
2x−1
x2 −2x+1
A
(x−1)
+
+
4
3
x+2
=1+
2x−1
(x−1)2
B
(x−1)2
• we get A = 2 and B = 1
• therefore r(x) =
Problem H 4.8
a)
1
2
b) ln(2)
c)
81
16
d) x2
2x−4
x2 −2x−8
=1+
2
(x−1)
+
1
(x−1)2