Sept 13, 2011
MATH 140: Calculus I
Tutorial 1
Some Noteworthy Functions
Exp(x)
Ln(x)
4
2
2
ln(x)
exp(x)
3
1
0
-1
0
-2
0
x
-4
1
0
x2
5
x
10
x3
100
1000
80
500
x3
x2
60
0
40
-500
20
0
-10
0
x
10
-1000
-10
0
x
10
Problem 1
Simplify b8 (2b)4 .
Solution:
b8 (2b)4 = b8 24 b4 = 16b(8+4) = 16b12
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Sept 13, 2011
MATH 140: Calculus I
Tutorial 1
Problem 2
Sketch f (x) = 1 − 12 e−x .
Solution:
1-1/2 e-x
2
0
f(x)
-2
-4
-6
-8
-10
-3
-2
-1
0
x
1
2
3
Problem 3
Sketch f (x) = e|x| .
Solution:
e|x|
60
50
f(x)
40
30
20
10
0
-4
-3
-2
-1
0
x
1
2
3
4
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Sept 13, 2011
MATH 140: Calculus I
Tutorial 1
Problem 4
Simplify the following: log5 (125).
Solution:
log5 (125) = log5 (5 · 5 · 5) = log5 (53 ) = 3.
Problem 5
Simplify the following: ln( 1e ).
Solution:
ln(1/e) = ln(e−1 ) = −1.
Problem 6
Simplify the following: e−2 ln 5 .
Solution:
e−2 ln 5 = eln 5
−2
= 5−2
Problem 7
Simplify the following: ln(a + b) + ln(a − b) − 2 ln(c).
Solution:
ln(a + b) + ln(a − b) − 2 ln(c) = ln((a + b)(a − b)) − 2 ln(c) = ln
(a + b)(a − b) c2
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Sept 13, 2011
MATH 140: Calculus I
Tutorial 1
Problem 8
Find the inverse: f (x) = 1 +
Solution:
√
2 + 3x.
y =1+
y−1=
√
√
2 + 3x
2 + 3x
2
(y − 1) = 2 + 3x
(y − 1)2 − 2 = 3x
(y − 1)2 − 2
=x
3
Therefore f −1 (y) =
(y−1)2 −2
.
3
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