Chapter 4—Suggested Solution
A. Sketch the graphs of the following functions, find the extrema and inflection points:
(i) f(x) = x ex
16
14
12
10
8
6
4
2
0
-2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
(ii) f(x) = ex / x
2500
2000
1500
1000
500
0
-500
-1000
-2
0
2
4
6
8
10
(iii) f(x) = x exp(-x2)
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
-0.1
-0.1
-0.2
-0.2
-0.3
-0.3
-0.4
-0.4
-0.5
-10
-0.5
-2
0
10
20
30
40
50
0
2
4
6
8
10
B. Given f(x) = ex + x2 - 2 , prove that
(i) f(x) has at least one root in the interval (0, 1), [ c is a root of f(x) if f(c)=0].
(ii) f(x) has only one root in the interval (0, 1).
Solution:
(i)f(x) has at least one root in the interval (0, 1), [ c is a root of f(x) if f(c)=0].
Solution:
f (0)  1, f (1)  0, f ( x) is continuous in the interval (0,1)
Hence : there is at least one c, s u c h t hfa(ct )  0
(ii) f(x) has only one root in the interval ( 0, 1)
, [hint: is f
monotonic in (0, 1)?].
Solution: f ( x)  e x  2 x  0 in the interval (0,1)
And f(x) is monotonic increasing in the interval (0,1); the value is from -1 to 0.
Hence, f(x) has only one root in the interval ( 0, 1)
§ 4.1 Exponential Function
Page 298
6, 18, 20, 34
6. Evaluate the given expressions.
a. (128) 3 / 7  8
 27 
b.  
 64 
2/3
 64 
 
 25 
3/ 2
2
3
32 * 9 288
 3 8
    

25
25
 4 5
18. Use the properties of exponents to simplify the given expressions.
a. ( x 2 y 3 z ) 3  x 6 y 9 z 3
 x 3 y 2
b.  4
 z



1/ 6
 x1 / 2 y 1 / 3 z 2 / 3
20. Find all real numbers x that satisfy the equation 3 x 2 2 x  144 .
Solution:
12 x  144  x  2
34. POPULATION GROWTH
It is estimated that the population of a certain
country grows exponentially. If the population was 60 million in 1997 and 90 million
in 2002, what will the population be in 2012?
Solution (e.g. 4.1.7)
Let P (x ) denote the population measured in million unit’s t years after the base
year 1997. Then P is a function of the form
P( x)  60e kt
Since the population was 90 million in 2002, it follows that
3
90  60e 5 k
Or e 5 k 
2
In 2012, the population will be
3
P(15)  60e15k  60(e 5 k ) 3  60( ) 3  202.5 million.
2
§ 4.2 Logarithmic Function
Page 314
6, 8, 20, 31, 36, 43
In Problems 6 and 8, evaluate the given expression using properties of the logarithm.
6. e 2 ln 3  9
8. ln
e3 e
 19 / 6
e1 / 3
4


x
20. Use logarithmic rules to simplify ln 

3
2
 x 1 x 
 4 x

11
1
2
ln 
   ln( x)  ln( 1  x )
3
2
4
2
 x 1 x 
In Problems 31 and 36, solve the given equation for x.
1
31. ln x  (ln 16  2 ln 2)
3
Solution (e.g. 4.2.8)
Using the logarithm rules we find that
1
1
(ln 16  2 ln 2)  (4 ln 2  2 ln 2)  2 ln 2  ln 4
3
3
That is , ln x  ln 4 . Take the natural logarithm of each side of this equation to get
e ln x  e ln 4
5
3
1  2e x
Solution (e.g. 4.2.8)
36.
Using the exponential rules we find that
5  3  6e  x
e x  3
x   ln 3
or
x4
x   ln 3 . Take the natural logarithm of each side of this equation to get
That is ,
43. COMPOUND INTEREST
How quickly will money double if it is invested at
an annual interest rate of 6% compounded continuously?
Solution (e.g. 4.2.10)
Apply the doubling time theory (page 311) to find that the money will be double
when
ln 2 ln 2
t

 11.55 years.
k
0.06
§ 4.3
Differentiation of Logarithmic and Exponential Function
Page 330
7, 36, 55
7. Differentiate the functions: (i) f ( x)  ( x 2  3x  5)e 6 x , (ii) ln(1+ x2) / [1 + exp(-x2)]
(iii) (ex –e-x) / (ex + e-x)
Solution (i) f ' ( x)  (2 x  3)e6 x  ( x 2  3x  5)e6 x  6
 (6 x 2  20 x  33)e6 x
(ii)
2x
 x2
2
 x2
(
1

e
)

ln(
1

x
)(
1

2
xe
)
2
1

x
f ' ( x) 
2
(1  e  x ) 2
(iii)
f ' ( x) 
(e x  e  x ) 2  (e x  e  x ) 2
4
 x
x
x 2
(e  e )
(e  e  x ) 2
36. Find the largest and smallest values of the function F ( x)  e x 2 x over interval
[0, 2].
Solution (Example 4.3.9)
Using the chain rule with u  x 2  2 x to find
2
F ' ( x)  e x 2 x  (2 x  2)  2( x  1)e x 2 x
2
so F ' ( x)  0
2
when
2( x  1)e x 2 x  0
x 1  0
x 1
Evaluating F(x) at the critical number x=1 and at the endpoints of the interval, x=0
and x=2, we find that
2
and
F (0)  e0  1
F (1)  e12  e 1  0.368
F (2)  e 44  1
Thus, F(x) has its largest value 1 at x=1 and x=2 and its smallest value 0.368 at x=1.
55. Given f ( x)  5 x , use the logarithmic differentiation to find the derivative
f ' ( x) .
2
Solution (e.g. 4.3.12, 4.3.13)
Use logarithm differentiation as follows
f ( x)  5 x
ln f ( x)  x 2 ln 5
2
f ' ( x)
 2 ln 5  x
f ( x)
f ' ( x)  2 ln 5  x  f ( x)
f ' ( x)  2 ln 5  x  5x
2
§ 4.4 Additional Exponential Models
Page342 25, 27, 38, 39
25. THE SPREAD OF AN EPIDEMIC
Public health records indicate that t
weeks after the outbreak of a certain form of influenza, approximately
2
f (t ) 
thousand people had caught the disease.
1  3e 0.8t
a. Sketch the graph of f (t )
b. How many people had the disease initially?
c. How many had caught the disease by the end of 3 weeks?
d. If the trend continues, approximately how many people in all will contract the
disease?
Solution (e.g. 4.4.5)
a. According the analysis in page 330, we find that the curve of the function f (t )
begins at f (0) 
point at t 
2
1
 , rises sharply (concave up) until it reaches the inflection
1 3 2
ln 3
, and then flatters out as tit continues to rise (concave down)
0 .8
toward the horizontal asymptote y  2 . We can sketch the graph of function f (t )
as follows.
b. Since f (0)  0.5 , there were 500 people
Y=2
had caught the disease initially.
c.
2
f (3) 
2
 1.572
1  3e 2.4
Thus , there were 1572 people had caught the
disease by the end of 3 weeks.
d. Since f (t ) approaches 2 as t increases without
ln3/0.8
bound, it follows that approximately 2000 people will contract the disease.
27. EFFICIENCY
The daily output of a worker who has been on the job for t
weeks is given by a function of the form Q(t )  40  Ae  kt . Initially the worker could
produce 20 units a day, and after 1 week the worker can produce 30 units a day. How
many units will the worker produce per day after 3 weeks?
Solution (e.g. 4.4.4)
Since the worker could produce 20 units a day initially and 30 units a day after 1
week, that is, Q(0)  20 and Q(1)  30 , we can get the equations
40  Ae k  30
and
40  A  20
Then solve A and k from the equations, A  20 and k  ln 2 . That is,
Q(t )  40  20e  ln 2t . After 3 weeks , the worker will produce
20
Q(3)  40  20e ln 23  40  3  37.5 units per week.
2
39. Newton's law of cooling states that the temperature of a hot object will cool down to the
temperature of its surrounding following an exponential decay. In symbols, if an object is
at a temperature T at time t and the surrounding is at a constant temperature
(e for
external), then T can be expressed as a function of time t:
T(t) = Te + (T0 – Te) exp(-kt).
If a piece of steak leaves an oven at a temperature of 200oC in a room at 20oC. It cools to
100oC in 15 minutes. How long does it take to cool further from 100 oC to 40oC?
Solution:
T (t )  T  (T0  T )e  kt
T (15)  20 0 C  (200 0 C  20 0 C )e  k15  100 0 C
20 0 C  (200 0 C  20 0 C )e kt  40 0 C
T  25.642
It takes to cool further from 100 oC to 40oC about dt= 25.6-15=10.6 minutes
§ 4.5 Differentiation of Trigonometric and Exponential Function
C. Differentiate the functions, where a & b are constants: (1) sin 2(ax) + cos 2(bx);
(2)ex sin(ax); (3) (sin(ax) + cos(bx)) exp(-x2) ; (4)cos(ax) / (ex + e-x).
Differentiate the functions
(1) sin 2(ax) + cos 2(bx)’= 2asin (ax)cos(ax) -2b cos (bx)sin(bx)
(2) ex sin(ax)’= ex (sin(ax)+acos(ax))
(3) ) (sin(ax) + cos(bx)) exp(-x2)’ =(acos(ax)-bsin(bx)) exp(-x2) -2x
exp(-x2) (sin(ax) + cos(bx))
(4) cos(ax) / (ex + e-x)=[-asin(ax) (ex + e-x)- (ex +-e-x)cos(ax)]/ (ex + e-x)2
§ 4.6 Taylor Expansion
D. Expand the following functions around x=0:
(i) ln(1+x),
(iii)
(iv)
Solution:
(ii) sin(x)
f ( x)  x 2  2 x 3
f ( x)  3xe x
2
/2
n
1 2 1 3
n 1 x
ln( 1  x)  x  x  x    (1)

2
3
n
x  (1,1]
1 3 1 5
(1) n 2 n 1
sin x  x  x  x   
x

3!
5!
(2n  1)!
x  (,);
f ( x)  x 2  2 x 3
Since:
1
1 2
1 3 3
(2n  3)!! n
1 x  1 x 
x 
x    (1) n
x   x [1,1]
2
24
246
(2n)!!
Hence:
 1
1
 2 x 2  1 3  2 x 3    (1) n (2n  3)!!  2 x n  
f ( x)  x 1  2 x  x1   2 x  
24
246
(2n)!!
 2


(2n  3)!! n1
 x  x 2   (1) 2 n1
x
 2 x  [1,1]
n!
n2
f ( x)  x 2  2 x 3
Since:
ex  1 x 
Hence:
f ( x)  3 xe
 x2 / 2
1 2
1
x    x n   x  (,);
2!
n!
n
   x2  1   x2 2

1   x2 
  
    
  
 3 x 1  
n!  2 
  2  2!  2 

n
  3x 3  3x 5
1  1 x 2 n1
 
 3 x  


n!
2n
 2  8
 x2
 (,);
2
 x2
 (,)
2
E. Use the Taylor Expansion to evaluate the following: (i) ln(1.5); (ii) sin(30o). How many
terms you need in order to achieve accuracy up to 3 decimal places.
Solution
ln( 1  x)  x 
1 2 1 3
xn
x  x    (1) n 1
  x  (1,1]
2
3
n
For ln(1.5), x=0.5, if 0.5n/n≤0.001, n≥8
(0.58/8=0.000488)
1 3 1 5
(1) n 2 n1
sin x  x  x  x   
x

3!
5!
(2n  1)!
x  (,);
for sin 30 0
x

6
,
x 2n1
 0.001, n  3
(2n  1)!