MHR • Calculus and Vectors 12 Solutions 477 Chapter 5 Exponential and Logarithmic Functions Chapter 5 Prerequisite Skills Chapter 5 Prerequisite Skills Question 1 Page 250 a) b) c) Answers may vary. For example: The equation of the inverse is 2 log y x = since 2 log 2 x = x . Chapter 5 Prerequisite Skills Question 2 Page 250 a) 2 x y = : Domain: (–∞, ∞) Range: (0, ∞) 2 log y x = : Domain: (0, ∞) Range: (–∞, ∞) b) 2 x y = : x-intercept: none y-intercept: 1 2 log y x = : x-intercept: 1 y-intercept: none c) 2 x y = : The function is increasing on the interval: x ! (–∞, ∞) 2 log y x = : The function is increasing on the interval: y ! (–∞, ∞) d) 2 x y = : y = 0 2 log y x = : x = 0
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MHR • Calculus and Vectors 12 Solutions 477
Chapter 5 Exponential and Logarithmic Functions Chapter 5 Prerequisite Skills
Chapter 5 Prerequisite Skills Question 1 Page 250
a)
b)
c) Answers may vary. For example:
The equation of the inverse is 2logy x= since 2log
2x
= x .
Chapter 5 Prerequisite Skills Question 2 Page 250 a) 2x
y = : Domain: (–∞, ∞)
Range: (0, ∞)
2logy x= : Domain: (0, ∞)
Range: (–∞, ∞)
b) 2xy = : x-intercept: none
y-intercept: 1
2logy x= : x-intercept: 1
y-intercept: none
c) 2xy = : The function is increasing on the interval: x ! (–∞, ∞)
2logy x= : The function is increasing on the interval: y ! (–∞, ∞)
d) 2xy = : y = 0
2logy x= : x = 0
MHR • Calculus and Vectors 12 Solutions 478
Chapter 5 Prerequisite Skills Question 3 Page 250
a) 32 = 8 b) 3.52 11.3=!
c) 1.52 2.8=!
d) 2log 10 3.3=!
e) 2log 7 2.8=!
f) 2log 4.5 2.2=!
Chapter 5 Prerequisite Skills Question 4 Page 250
a) 32 = 8 b) 3.52 11.3=!
c) 1.52 2.8=!
d) 2log 10 3.3=!
e) 2log 7 2.8=!
f) 2log 4.5 2.2=!
Chapter 5 Prerequisite Skills Question 5 Page 250 a) 3
After half an hour, the amount remaining is 1.5625 g.
ii) Half a day is 12 × 60 min = 720 min.
5
–
7 0
4
2
2
1( ) 100
2
4.484 1
720
0
A! "
= # $% &
= '!
After half a day, the amount remaining is 4.484! 10
–42 g.
MHR • Calculus and Vectors 12 Solutions 485
Chapter 5 Section 1 Rates of Change and the Number e
Chapter 5 Section 1 Question 1 Page 256 a)
b)
c)
d)
MHR • Calculus and Vectors 12 Solutions 486
Chapter 5 Section 1 Question 2 Page 256 a)
b)
c)
d)
Chapter 5 Section 1 Question 3 Page 256
a) ( ) 2xf x = :
{x !!}
( ) xf x e= :
{x !!}
b) No
c) No
MHR • Calculus and Vectors 12 Solutions 487
Chapter 5 Section 1 Question 4 Page 257
a) B. The graph of the derivative of a quadratic function is a straight line.
b) C. The graph of the derivative of a line is of the form y = a, where a is a constant.
c) D. The graph of the derivative of an exponential function is also an exponential function.
d) A. The graph of the derivative of a cubic function is a quadratic function.
Chapter 5 Section 1 Question 5 Page 257 a) b > e
b) 0 ≤ b < e
Chapter 5 Section 1 Question 6 Page 258 a)
b) Answers may vary. For example:
The graph of the rate of change of 1
2
x
y! "
= # $% &
will be a compression and a reflection of the graph of
y in the x-axis.
c)
MHR • Calculus and Vectors 12 Solutions 488
Chapter 5 Section 1 Question 7 Page 258
Answers may vary. For example:
If 0 < b < 1, the graph of x
y b= will be above the x-axis and the graph of the rate of change of this
function will be below the x-axis. If b > 1, the graph of x
y b= and the graph of the rate of change of
this function will both be above the x-axis.
Chapter 5 Section 1 Question 8 Page 258 a)
b)
c) Answers may vary. For example:
The graph of the combined function ( )g x will be a horizontal straight line.
MHR • Calculus and Vectors 12 Solutions 489
d)
Answers may vary. For example:
The graph of ( ) ln 4g x = is a constant function. Therefore the graph is a horizontal straight line.
Chapter 5 Section 1 Question 9 Page 258 a) Answers may vary. For example:
No. The shape of the graph of g will not change. The shape of the graph of g will be a horizontal
straight line. If the base is other than 4, the graph will be parallel to the graph of g and shifted up or
down depending on the numerical value of the base.
If the value of the base is greater than 4, the graph will be shifted up. If the value of the base is
greater than 1 and less than 4, the graph will be shifted down, but will still be above the x-axis.
If the value of the base is greater than 0 and less than 1, the graph will be shifted down and will be
below the x-axis.
b) The graph is the line ( ) lng x e= , which is the horizontal straight line ( )
( )( )
f xg x
f x
!= where ( ) 1g x = .
Chapter 5 Section 1 Question 10 Page 258 Solutions to the Achievement Checks are shown in the Teacher’s Resource. Chapter 5 Section 1 Question 11 Page 258 a) Answers may vary. For example:
The graph of the function ( )g x! will be the horizontal straight line, y = 0.
b) Answers may vary. For example:
If ( ) 2xf x = , then ( ) 2 ln 2x
f x! = , and 2 ln 2
( )2
x
xg x = .
( ) ln 2g x = , which is just a constant so ( )g x! = 0. This is applicable for any base. Therefore, the
graph of ( )g x! will be the graph of y = 0.
The graph of ( )g x is a constant function. The derivative of a constant function is 0.
MHR • Calculus and Vectors 12 Solutions 490
c) Answers may vary. For example:
The function ( )g x will be a horizontal straight line for any value of b, b > 0 and the derivative
function, ( )g x! , will be ( ) 0g x! = , for any value of b, b > 0, since the derivative of any constant
function is the horizontal straight line y = 0.
Chapter 5 Section 1 Question 12 Page 258
a) {x !!}
b) {y < 0 < 1, y !!}
c) As the value of c increases, the graph of the function is shifted to the right.
Chapter 5 Section 1 Question 13 Page 258
Answers may vary. For example:
Some answers include: Leonhard Euler; 1727; e is used in probability
Chapter 5 Section 1 Question 14 Page 258
B
Chapter 5 Section 1 Question 15 Page 258
D
MHR • Calculus and Vectors 12 Solutions 491
Chapter 5 Section 2 The Natural Logarithm
Chapter 5 Section 2 Question 1 Page 265 a)
b) i) Domain:
{x !!}
ii) Range: {y < 0, y !!}
iii) x-intercepts: none; y-intercept: –1
iv) Horizontal asymptote: y = 0
v) Decreasing on the interval (–∞, ∞)
vi) Maximum or minimum points: none
vii) Points of inflection: none
Chapter 5 Section 2 Question 2 Page 265 a)
b) i) Domain:
{x > 0, x !!}
ii) Range: {y !!}
iii) x-intercept: 1; y-intercepts: none
iv) Vertical asymptote: x = 0
v) Decreasing on the interval (0, ∞)
MHR • Calculus and Vectors 12 Solutions 492
vi) Maximum or minimum points: none
vii) Points of inflection: none
Chapter 5 Section 2 Question 3 Page 265
Answers may vary. For example:
No. The function ( )f x and the function ( )g x are not inverse functions. They are not reflections of
each other in the line y = x.
Chapter 5 Section 2 Question 4 Page 265 Answers may vary. For example:
The value of ln 0 is –∞, which is undefined. Also, the domain of the function lny x= is the interval
(0, ∞), so when x = 0 the function is undefined.
MHR • Calculus and Vectors 12 Solutions 493
Chapter 5 Section 2 Question 8 Page 265
a) 2ln( ) 2 ln
2
xe x e
x
=
=
b) ln( ) ln( ) 2ln( )
2 ln
2
x x xe e e
x e
x
+ =
=
=
c) ln( 1) 1xe x
+= +
d) ln(3 ) 2
2
(ln( )) 3 (2 ln )
6
x xe e x x e
x
=
=
Chapter 5 Section 2 Question 9 Page 265
a) 5
ln ln5
1.609
x
x
e
e
x
=
=
=!
b) 4
4
4
1000 20
50
ln ln50
4ln50
15.648
x
x
x
e
e
e
x
x
=
=
=
=
=!
c) ln( ) 0.442
0.442
xe
x
=
=
d) ln(2 )7.316
2 7.316
3.658
xe
x
x
=
=
=
Chapter 5 Section 2 Question 10 Page 265 a) 3 15
ln3 ln15
ln15
ln3
2.465
x
x
x
x
=
=
=
=!
MHR • Calculus and Vectors 12 Solutions 494
b) 3 15
log3 log15
log15
log3
2.465
x
x
x
x
=
=
=
=!
c) Answers may vary. For example:
The value of x can be found by taking natural logarithms of both sides of the equation or by taking
common logarithms of both sides of the equation.
Chapter 5 Section 2 Question 11 Page 265
a) 4max
max 4max
4
( )
=2
1ln ln
2
14ln
2
2.8
t
t
t
V t V e
VV e
e
t
t
!
!
!
=
" # " #=$ % $ %
& '& '" #
= ! $ %& '
=!
It will take 2.8 s.
b) 4max
max 4max
4
( )
= 10
1ln ln
10
14ln
10
9.2
t
t
t
V t V e
VV e
e
t
t
!
!
!
=
" # " #=$ % $ %
& '& '" #
= ! $ %& '
=!
It will take 9.2 s.
MHR • Calculus and Vectors 12 Solutions 495
Chapter 5 Section 2 Question 12 Page 266 a)
Use the ExpReg function:
The equation of the function is:
0.86
( )
7
200
200( )
t
k
t
T t e!
=
=
Taking the logarithm of both sides,
ln ln(0.867)
ln(0.867)
1
ln(0.867)
7
t
tke
tt
k
k
k
!
=
! =
= !
=
b)
T (10) = 200e!
10
7
!= 48
T (10) = 200(0.867)10
!= 48
At 10 min, the temperature is 48ºC.
c)
T (15) = 200e!
15
7
!= 23
At 15 min, the temperature is 23ºC.
Answers may vary. For example:
The pizza will reach room temperature (21ºC) after a long time.
MHR • Calculus and Vectors 12 Solutions 496
Chapter 5 Section 2 Question 13 Page 266
a) ln 2 ln3 1.7918+ =!
b) ln 6 1.7918=!
c) Answers may vary. For example:
The results seem to verify the Law of Logarithms for Multiplication. In terms of natural
logarithms, the Law of Logarithms for Multiplication of natural logarithms is
ln( ) ln lna b a b! = + , a > 0, b > 0.
Chapter 5 Section 2 Question 14 Page 266
a) i) (ln 2)
0 57000
(ln 2)
5700
10
ln ln 0.1
5700ln 0.1
ln 2
18 935
t
t
NN e
e
t
t
!
!
=
" #=$ %
& '
= !
=!
The age is 18 935 years.
ii) (ln 2)
0 57000
(ln 2)
5700
100
ln ln 0.01
5700ln 0.01
ln 2
37 870
t
t
NN e
e
t
t
!
!
=
" #=$ %
& '
= !
=!
The age is 37 870 years.
iii) (ln 2)
0 57000
(ln 2)
5700
2
ln ln 0.5
5700ln 0.5
ln 2
5700
t
t
NN e
e
t
t
!
!
=
" #=$ %
& '
= !
=
The age is 5700 years.
b) Answers may vary. For example:
No. The half-life of C-14 is approximately 5700 years. It will take 5700 years for the sample to
have a C-14 to C-12 ratio of half of today’s level and it will take 11 400 years for the sample to
have a C-14 to C-12 ratio of one quarter of today’s level.
MHR • Calculus and Vectors 12 Solutions 497
c)
( )
0
5700 ln
ln 2
N t
Nt
! "# $% &
= '
Chapter 5 Section 2 Question 15 Page 266 a)
b) Answers may vary. For example:
The shape of the graph is similar to the shape of the normal distribution curve.
c) The maximum value is y = 1 and this occurs when x = 0.
d) Answers may vary. For example:
From the graph, use a trapezoid to estimate the area with a base of 3 units, top 0.5 units, and height
1 unit.
A = ha + b
2
!
"#$
%&
= 13+ 0.5
2
!
"#$
%&
!= 1.75
This gives an estimate of 1.75 units2.
Note: The total area is given by the integral of the error function and is ! = 1.77 square units.
MHR • Calculus and Vectors 12 Solutions 498
e)
From the graph it can be seen that an estimate of the area between x = –1 and x = +1 can be made
by using the sum of the area of a rectangle and trapezoid.
area of rectangle = 2 ! 0.367
= 0.734
area of trapezoid = (1! 0.376)2 + 0.5
2
"
#$%
&'
= 0.78
Therefore, the estimated area between x = –1 and x = 1 is 0.734 units2 + 0.78 units
2 != 1.5 units
2.
Note: The Empirical Rule for normal distributions states that this area should be 68% of the total
area under the curve.
Chapter 5 Section 2 Question 16 Page 266
D
Chapter 5 Section 2 Question 17 Page 266 C
MHR • Calculus and Vectors 12 Solutions 499
Chapter 5 Section 3 Derivatives of Exponential Functions
Chapter 5 Section 3 Question 1 Page 274 a)
!g (x) = 4x ln 4 b)
!f (x) = 11x ln11
c)
dy
dx=
1
2
!
"#$
%&
x
ln1
2
d) !N (x) = "3e
x
e) ( ) xh x e! =
f)
dy
dx= !
x ln!
Chapter 5 Section 3 Question 2 Page 274
a) ( ) xf x e! = ;
!!f (x) = ex ;
!!!f (x) = ex
b) f
(n) (x) = ex
Chapter 5 Section 3 Question 3 Page 274
dy
dx= 5x ln5
dy
dxx=2
= 52 ln5
!= 40.2
The instantaneous rate of change is 40.2.
Chapter 5 Section 3 Question 4 Page 274
dy
dx=
1
2e
x
dy
dxx=4
=1
2e
4
!= 27.3
The slope is 27.3.
MHR • Calculus and Vectors 12 Solutions 500
Chapter 5 Section 3 Question 5 Page 274
dy
dx= 8x ln8
dy
dxx=
1
2
= 81
2 ln8
= 2 2( )3ln 2
= 6 2 ln 2
When x =1
2, y = 2 2
Substitute x1
=1
2, y
1= 2 2, and m = 6 2 ln 2 in y ! y
1= m(x ! x
1).
Therefore,
y ! 2 2 = 6 2 ln 2 x !1
2
"
#$%
&'
y = 6 2 ln 2( ) x + 2(2 ! 3ln 2)
Chapter 5 Section 3 Question 6 Page 274 a)
N (t) = 10(2t ) ; t is the time in days; N (t) is the number of fruit flies
b) 7( ) 10(2
2 0
7 )
1 8
N =
=
After 7 days, there will be 1280 fruit flies.
c)
Rate of increase = !N (t)
= 10(2t ) ln 2
!N (7) = 10(27 ) ln 2
!= 887
At 7 days, the rate is 887 fruit flies per day.
d)
500 = 10(2t )
ln50 = t ln 2
t =ln50
ln 2
t != 5.64
It will take 5.64 days for the population to reach 500 flies.
MHR • Calculus and Vectors 12 Solutions 501
e)
!N (5.64) = 10(25.64 ) ln 2
!= 346
At 5.64 days, the rate is 346 fruit flies per day.
Chapter 5 Section 3 Question 7 Page 275
a) i)
20 = 10(2t ) ln 2
2t=
2
ln 2
t =
ln2
ln 2
!"#
$%&
ln 2
t != 1.53
The time is 1.53 days.
ii)
2000 = 10(2t ) ln 2
2t=
200
ln 2
t =
ln200
ln 2
!"#
$%&
ln 2
t != 8.17
The time is 8.17 days.
b) Answers may vary. For example:
Since the growth rate increases exponentially, it is most desirable to begin an extermination
program very soon after 2 days. At 8 days, the population becomes out of control.
MHR • Calculus and Vectors 12 Solutions 502
Chapter 5 Section 3 Question 8 Page 275
!f (x) =1
2e
x
!f (ln3) =1
2e
ln3
=3
2
Therefore, the slope of the perpendicular line is –2
3.
When x = ln 3, y = 3
2.
1 1 1 1
3 2Substitute ln3, , and – in ( – ).
2 3x y m y y m x x= = = ! =
3 2– ln3( )
2 2 3– ln3
3 3 2
2 3y x
y x
! = !
= + +
Chapter 5 Section 3 Question 9 Page 275
y = !2
3x +
2
3ln3+
3
2
y != !2
3x + 2.2341
Chapter 5 Section 3 Question 10 Page 275
a) Answers may vary. For example:
The shape of the graph of ( )g x is a horizontal straight line.
b) !f (x) = kb
x ln b ;
g(x) =kb
x ln b
kbx
= ln b
c) The simplified form of the function is the graph of a horizontal straight line: ( ) lng x b= .
MHR • Calculus and Vectors 12 Solutions 503
Chapter 5 Section 3 Question 11 Page 275
Answers may vary. For example:
Use a graphing calculator. Let k = 5 and b = 3. Then
g(x) = ln3
!= 1.0986
Chapter 5 Section 3 Question 12 Page 275
a) ( ) 1g x = ; Answers may vary. For example:
The derivative of the exponential function of the form ( ) xf x ke= is ( ) x
f x ke! = .
The simplified form of the function ( )g x is the function
( )( )
( )
1
x
x
f xg x
f x
ke
ke
!=
=
=
b) ( ) 1g x =
Chapter 5 Section 3 Question 13 Page 275
a) f
(n) (x) = bx (ln b)n
b) Answers may vary. For example:
!f (x) = bx ln b, !!f (x) = b
x (ln b)2 , !!!f (x) = bx (ln b)3, ... , f (n) (x) = b
x (ln b)n
MHR • Calculus and Vectors 12 Solutions 504
Chapter 5 Section 3 Question 14 Page 275 a) Answers may vary. For example:
Both functions have the same y-value of 16, when the x-value is 4. Both of the functions are
increasing functions that do not have a local maximum or minimum point, or a point of inflection.
The function g(x) = 2x is increasing more rapidly than the function
2( )f x x= over the given
interval, 4 ≤ x ≤ 16.
2( )f x x= , 4 ≤ x ≤ 16 ( ) 2x
g x = , 4 ≤ x ≤ 16
b) Answers may vary. For example:
No. The derivatives of the two functions will not be similar. The derivative of the quadratic
function 2( )f x x= is ( ) 2f x x! = . When graphed, the derivative function is a linear function with a
slope of 2. The derivative of the exponential function ( ) 2xg x = is
!g (x) = 2x ln 2 . When graphed,
the derivative function is also an exponential function.
c) ( ) 2f x x! = , 4 ≤ x ≤ 16 !g (x) = 2x ln 2 , 4 ≤ x ≤ 16
d) Answers may vary. For example:
Yes. There are two x-values for which the slope of ( )f x will be approximately the same as the
slope of ( )g x when rounded to five decimal places. When the x-value is 0.485 09, the slope of
( )f x and the slope of ( )g x is 0.970 18. When the x-value is 3.212 43, the slope of ( )f x and the
slope of ( )g x is 6.424 87. The x-values can be found using a graphing calculator.
MHR • Calculus and Vectors 12 Solutions 505
Chapter 5 Section 3 Question 15 Page 275
a)
P(h) = 101.3e!kh
95.6 = 101.3e!1000k
ln95.6
101.3
"
#$%
&'= (!1000k) ln e
k = !0.001ln95.6
101.3
"
#$%
&'
k != 0.000 057 9
b)
P(2000) = 101.3e!(0.000 057 9)(2000)
!= 90.2
The pressure is 90.2 kPa.
c)
!P (h) = 101.3("0.000 057 9)e"0.000 057 9h
!= "0.00587e"0.000 057 9h
d)
!P (1500) = "0.005 87e"0.000 057 9(1500)
!= –0.005 38
The rate is −0.005 38 kPa/m.
Chapter 5 Section 3 Question 16 Page 276
a) ( ) 50(2 )tN t = where N is the number of visitors and t is the time in weeks.
b) i) 4( ) 50(2 )
8
4
00
N =
=
After 4 weeks, there will be 800 visitors.
ii) 12( ) 5 )1 0(22N =
After 12 weeks, there will be 204 800 visitors.
c)
!N (t) = 50(2t ) ln 2
i)
!N (4) = 50(24 ) ln 2
!= 555
The rate is 555 visitors per week.
ii)
!N (12) = 50(212 ) ln 2
!= 141957
The rate is 141 957 visitors per week.
d) Answers may vary. For example:
No. This trend will not continue indefinitely. The number of people visiting the site will eventually
level off.
MHR • Calculus and Vectors 12 Solutions 506
Chapter 5 Section 3 Question 17 Page 276 a) Answers may vary. For example:
One Internet site claims the current population is growing at a rate of 205 000 per day, or
8500 per hour, or 140 per minute, or 2.3 people per second.
b) Answers may vary. For example:
i) Equation form: P(t) = 4e
0.019t , where P is the population, in billions, and t is the time, in years,
since 1975. ii) Graphical form:
c) i)
P(50) = 4e0.019(50)
= 10.342 838 64
The population would be 10.342 838 64 billion.
ii)
P(525) = 4e0.019(525)
!= 85 930.521 67
The population would be 85 930.521 67 billion. iii)
P(1025) = 4e0.019(1025)
!= 1 148 008 296
The population would be 1 148 008 296 billion.
d) Answers may vary. For example:
No. This model is not sustainable over the long term. Other factors that could affect this trend are
the amount of resources available to sustain the population and the available areas on the earth that
could sustain this number of people.
e) Answers may vary. For example:
If the resources, such as food, start to diminish then the population increase would slow down,
since the death rate would increase relative to birth. Poor nutrition is one contributing factor to low
birth rates. The factor 0.019 would be reduced.
MHR • Calculus and Vectors 12 Solutions 507
Chapter 5 Section 3 Question 18 Page 276 a) Answers may vary. For example:
i)
10!6= 4e
0.019t
ln(2.5"10!7 ) = 0.019t
t != !800
Since t = 0 in 1975, the year when the population would have been 1000 is predicted
as 1975 – 800 = 1175.
ii)
10!7= 4e
0.019t
ln(2.5"10!8 ) = 0.019t
t != !921
Since t = 0 in 1975, the year when the population would have been 100 is predicted
as 1975 – 921 = 1054.
iii)
2 !10"9= 4e
0.019t
ln(5!10"10 ) = 0.019t
t != "1127
Since t = 0 in 1975, the year when the population would have been 2 is predicted
as 1975 – 1127 = 848.
b) Answers may vary. For example:
No. The answers in part a) do not seem reasonable.
Chapter 5 Section 3 Question 19 Page 276 Answers may vary. For example:
Students may use a graphing calculator to graph lny x= and use the tangent function for some values
of x.
x = 1: tangent is y = x !1 where the slope is 1
x = 2 : tangent is y = 0.5x ! 0.31 where the slope is 0.5
x = 0.5: tangent is y = 2x !1.69 where the slope is 2
Graph the derivative function of lny x= (i.e., 1( )f x x
!= ) and compare the ordered pairs with the
(x, slope) values for lny x= . They are the same:
MHR • Calculus and Vectors 12 Solutions 508
Chapter 5 Section 3 Question 20 Page 276
D
Chapter 5 Section 3 Question 21 Page 276 E
MHR • Calculus and Vectors 12 Solutions 509
Chapter 5 Section 4 Differentiation Rules for Exponential Functions Chapter 5 Section 4 Question 1 Page 282 a) lnx b
y e=
b)
dy
dx= e
x lnb ln b
Chapter 5 Section 4 Question 2 Page 282
a) 33 xdye
dx
!= !
b) 4 5( ) 4 xf x e
!" =
c)
dy
dx= 2e
2x! (!2)e!2x
= 2e2x
+ 2e!2x
d)
dy
dx= 2x ln 2 + 3x ln3
e)
f (x) = 3e2x! (2x )3
"f (x) = (2)3e2x! 3(2x )2 2x ln 2
= 6e2x! 3(23x ) ln 2
f)
dy
dx= 4xe
x+ 4e
x
= 4ex (x +1)
g)
dy
dx= 5x
e! x ln5! 5x (e! x )
= !5xe! x (1! ln5)
h) 2 2 3( ) 2 6x x x
f x xe e e!
" = + !
MHR • Calculus and Vectors 12 Solutions 510
Chapter 5 Section 4 Question 3 Page 282
a)
dy
dx= !e
! x sin x + e! x cos x
= e! x (cos x ! sin x)
b)
dy
dx= ! sin x(ecos x )
c)
!f (x) = 2e2x (x
2" 3x + 2) + e
2x (2x " 3)
= e2x (2x
2" 4x +1)
d)
!g (x) = 4xecos2x
+ 2x2e
cos2x ("2sin x)
= "4xecos2x (x sin 2x "1)
Chapter 5 Section 4 Question 4 Page 282
2
2
( ) 2
If ( ) 0 then 2 0.
x x
x x
f x e e
f x e e
! = "
! = " =
Therefore,
ex (1! 2e
x ) = 0
ex
=1
2 since ex
> 0
x = ln0.5
f (ln0.5) = eln0.5
! e2ln0.5
= 0.5! 0.52
= 0.25
Therefore, by using derivative tests, there is a local maximum of y = 0.25 when x = ln(0.5). Chapter 5 Section 4 Question 5 Page 282 Adding two exponential functions gives an exponential function.
!f (x) = ex+ 2e
2x
Set !f (x) = 0.
ex (1+ 2e
x ) = 0
ex must equal "
1
2 or 0 but since ex
> 0, the function has no local extrema.
MHR • Calculus and Vectors 12 Solutions 511
Chapter 5 Section 4 Question 6 Page 282 Graph of
2x xy e e= ! Graph of
2x xy e e= +
Chapter 5 Section 4 Question 7 Page 283
a)
P(3) = 50e0.5(3)
!= 224
After 3 days, there will be 224 bacteria.
b) Initial population = 50 (i.e., t = 0)
10(50) = 50e0.5t
10 = e0.5t
ln10 = 0.5t
t =ln10
0.5
t != 4.6
The time is 4.6 days.
c)
e0.5t
= 10x
0.5t = x ln10
x =t
2 ln10
x !=t
4.6
P(t) = 50e0.5t
!= 50(10)
t
4.6
d)
P(5) = 50(10)
5
4.6
!= 611
After 5 days, there will be 611 bacteria.
MHR • Calculus and Vectors 12 Solutions 512
e)
P(5) = 50e0.5(5)
!= 609
After 5 days, there will be 609 bacteria.
Answers may vary. For example:
The function from part c) approximates the relationship between e and t from the initial function.