Math 1205 – Worksheet 2 – Solutions
Problem 1:
Part (a):
»format long
»sqrt(3)
ans =
1.73205080756888
This is not the exact value for square root of 3 because it is a terminating decimal.
Part (b):
»x1 = [1.7, 1.73, 1.732, 1.7320, 1.73205, 1.732050];
»x2 = [1.8, 1.74, 1.733, 1.7321, 1.73206, 1.732051];
»[x1', (2.^x1)', x2', (2.^x2)']
ans =
1.70000000000000 3.24900958542494 1.80000000000000
1.73000000000000 3.31727818325777 1.74000000000000
1.73200000000000 3.32188009636358 1.73300000000000
1.73200000000000 3.32188009636358 1.73210000000000
1.73205000000000 3.32199522594976 1.73206000000000
1.73205000000000 3.32199522594976 1.73205100000000
(i)
x1 is smaller and x2 is larger than the exact value of
(ii)
2 x1  2
3
3.48220225318450
3.34035167771348
3.32418344637459
3.32211035952609
3.32201825234581
3.32199752858218
3.
 2x2
(iii)
21.732050 and 21.732051 are closer to the value of 2
accurate approximations to 3
3
because the exponents are more
(iv)
3
The average of two values closest to 2 gives
»(3.32199522594976 + 3.3219975285821)/2
ans =
3.32199637726593
Part (c)
2 3  lim 2 x for x a rational number
x 3
Problem 2:
Part (a):
»x = [53.93/52.83, 55.07/53.93, 56.23/55.07, 57.41/56.23, 58.61/57.41, ...
59.84/58.61, 61.10/59.84, 62.38/61.10, 63.69/62.38];
»[x']
ans =
1.02082150293394
1.02113851288708
1.02106410023606
1.02098523919616
1.02090228183243
1.02098617983279
1.02105614973262
1.02094926350246
1.02100032061558
a=1.021
Part (b)
P(1)=a*P(0)
P(2)=a*P(1)=a^2*P(0)
P(3)=a*P(2)=a^2*P(1)=a^3*P(0)
P(t)=a^t*P(0) or P(t) = 52.83(1.021)t.
Part (c)
»P11 = 52.83*(1.021)^11
P11 =
66.39934169385280
Approximately 66.4 million people.
Part (d)
Find t when P(t)=90
52.83(1.021)^t=90
(1.021)^t=1.704
t*ln(1.021)=ln(1.704)
t = ln(1.704)/ln(1.021) = 25.64549
According to this model, 26 years after 1990 (in the year 2016) the population would
reach 90 million.
Problem 3:
Part (a)
»n = [100, 200, 300, 400, 500];
»y = (1 + 1./n).^n;
»[n', y']
100.0000 2.7048
200.0000 2.7115
300.0000 2.7138
400.0000 2.7149
500.0
2.7156
e is approximately 2.72
Part (b) (One of many possible tables)
»n=5000:1000:10000;
»y=(1+1./n).^n;
»[n',y']
ans =
1.0e+04 *
0.50000000000000 0.00027180100501
0.60000000000000 0.00027180553396
0.70000000000000 0.00027180876909
0.80000000000000 0.00027181119553
0.90000000000000 0.00027181308282
1.00000000000000 0.00027181459268
e is approximately 2.718
Part (c)
(i)
»[exp(2),exp(-2),exp(sqrt(2))]
ans =
7.38905609893065 0.13533528323661 4.11325037878293
(ii)
»[2.718^2,2.718^(-2),2.718^(sqrt(2))]
ans =
7.38752400000000 0.13536335042702 4.11264729009486
Problem 4:
Part (a)
»x = -2:.01:2;
»plot(x,exp(x),x,2.^x,x,3.^x)
10
8
6
4
2
0
-2
-1
(i) For x>0, 2 x  ex  3 x
(ii) For x<0, 3x  ex  2 x
0
1
2
Part (b)
»x = 0:.01:2;
»plot(x,5.^x,x,x.^5)
35
30
25
20
15
10
5
0
0
1
2
»x = 2:.01:6;
»plot(x,5.^x,x,x.^5)
2
x 10
4
1.5
1
0.5
0
2
4
6
»x = 6:.01:8;
»plot(x,5.^x,x,x.^5)
4
x 10
5
3
2
1
0
6
7
8
(i) The graphs intersect near x =1.76 and at x = 5.
x
5
(ii) 5  x when 0<x<1.76 and when 5<x<8
x
5
(iii) 5  x when 1.76<x<5
Part (c) (i)
»x=.1:.01:10;
»plot(x,log(x))
3
2
1
0
-1
-2
-3
0
2
4
6
8
10
(ii) »zoom on
1.02
1.01
1
0.99
0.98
2.68
2.7
2.72
2.74
log(2.72) is approximately 1
(ii) The base being used is e (which is approximately 2.72).
Let a be the base. Then 2.72  log(1) = a1 = a.