Solutions to Spring 2014 BC 1 Semester Review
1a.
1b.
1c.
1d.
1e.
1f.
1g.
1h.
1i.
2.
6
3x  6x
x  lim
x2  3 0  3
lim

1
2
x  1  2x 3
x  1
2 0 2
x3
2
x
1  cos 3x 
1  cos 3x  3
lim
 lim
  0 3  0
x
3x
1
x 0
x 0
sin 3x 
sin 3x  cos 2x 
sin 3x 
cos 2x  3
2x
3 3
lim
 lim

 lim


  111 
1
sin 2x  x  0 3x
sin 2x 
1
2
2 2
x  0 tan 2x 
x 0
1
3
3
3
4
5
 3
4x  5x
00
lim
 x  lim x x 
0
4
1
1
2

0
x  2x  1
x 
4
2 4
x
x
x 5
x 5
lim
 lim 
 1
x 5 5  x
x 5 x  5
 x  4   x  3  lim  x  4   3  4   7
x 2  x  12
lim
 lim
2x  6
2  x  3
2
2
2
x 3
x 3
x 3
x 4
lim
 
x 2 2  x
x 3
x 3
lim 2
 lim

x 2 x  2x
x 2 x  x  2
1
3
lim
 0
sin3  4 
3
4
3
sin3  4  64
 sin  4  
3
 lim


64
lim

  64  1  64
3
1
4 
 0 64
 0 
Only point where f might not be continuous is at x = 1.
(1)
Does f(1) exist? Yes, f(1) = 2
(2)
Does lim f (x ) exist? Yes
x 1
lim f (x )  lim x 2  3  12  3  2
x 1
x 1
lim f (x )  lim 2x  4  2  4  2
x 1
Since lim f (x )  lim f  x  , lim f (x ) exists.
x 1
(3)
x 1
x 1
x 1
Does lim f (x )  f 1  . No, since lim f (x )  2  2  f 1  .
x 1
x 1
Since f is not continuous is at x = 1, f is not continuous.
Spring 2014 BC 1 Semester Review Solutions p. 1
SP 14
3.
Evaluating f(1) = 14 – 6 + 2 = -3, and f(2) = 24 – 12 + 2 = 6, we see that f(1) < 0, f(2) > 0.
Since f is continuous on the closed interval [1, 2], it satisfies the hypotheses of the IVT,
which then guarantees that f hits every value between –3 and 6, including 0.
4.
Determine values of x near x = 2 for which x 3  3  5  0.1 or x 3  8  0.1 .


x 3  8  0.1  0.1  x 3  8  0.1  7.9  x 3  8.1  1.992  x  2.008


So, if x  1.992,2.008 , then x 3  3  5  0.1 .
5.
8
6
4
2
1.0
0.5
0.5
1.0
1.5
2.0
Determine endpoints of segments that form approximation for y, starting with y(–1) = –5.
First step: 1  x  0
Second step: 0  x  1
Third step: 1  x  2
ynew  yold  m  x
ynew  yold  m  x
ynew  yold  m  x
 5  8  1
3
 3  31
6
Spring 2014 BC 1 Semester Review Solutions p. 2
 6  0 1
6
SP 14
6
(1, 6)
(2, 6)
4
2
1.0
(0, 3)
0.5
0.5
1.0
1.5
2.0
2
4
(–1, –5)
f a  h   f a 
6.
f   a   lim
7a.
f  x   x 2  3x  1
h 0
h
OR f  a   lim
x a
f  x   f a 
x a
2

 x  h   3  x  h   1  x 2  3x  1
f   x   lim
h
h 0
x 2  2xh  h 2  3x  3h  1  x 2  3x  1
 

 lim 
h
h 0
2
2xh  h  3h
 lim
h
h 0
 lim 2x  h  3
h 0
 2x  3
7b.
g x   x  2
Spring 2014 BC 1 Semester Review Solutions p. 3
SP 14
g   x   lim
h 0
 lim
h 0 h
 lim
h 0 h
 lim
h 0

7c.
h x  
x h 2  x 2 x h 2  x 2

h
x h 2  x 2
 x  h  2   x  2
 x h 2  x 2
h
 x h 2  x 2
1
x h 2  x 2
1
2 x 2
1
x 2
1
1

h   x   lim x  h  2 x  2
h
h 0
1  x  2   x  h  2
 lim 
h 0 h
 x  h  2 x  2
1
h

h 0 h  x  h  2  x  2
 lim
1
h 0  x  h  2  x  2
 lim

8a.
1
x
2
 2
2
1 2
x  3x  4 3  2x  3 .
3
f is not differentiable at values of x for which f  is undefined, which occur when:
x 2  3x  4  0

f  x   3 x 2  3x  4  x 2  3x  4
x

1
3
 f  x  


 4   x  1  0
x  4 or x  1
8b.
g  x   x 2  3x  10 is not differentiable at values of x for which g  x   0 , since at
these values of x, the graph of g will have corners.
Spring 2014 BC 1 Semester Review Solutions p. 4
SP 14
x 2  3x  10  0
x
 5  x  2  0
x  5 or x  2
9.
Many answers are possible. The three types, with one example of each, are given below.
Corner
Cusp
y  x 2
y   x  2
Vertical Tangent
2
3
y   x  2
1
3
2
1.5
1
1.25
1.5
0.5
1
1
0.75
1
0.5
0.5
2
3
4
-0.5
0.25
-1
1
10a.
10b.
2
3
4
1
2
3
4
 2
  0

 3  
 3

s 2   s  0   2  1
 0 1
1
Average velocity over first 2 seconds:

20
2
3
Instantaneous velocity at t = 2 seconds is s  2 .
s  t  
1  t  1   t  1 
t
1
2
 1

1
t
2
 1
 s  2  
10c.
Since s  t  
11.
Use Nderiv on the calculator, f  1   0.173 .
t
Can also calculate
h  0.01 :
2
 1
h
f 1  h   f 1 
h  0.001 :
 sin 1.01

f 1  h   f 1 
h  0.0001 :
0.01
 sin 1.001

f 1  h   f 1 
1
0.001
0.001
h
2  1 

1
9
for small values of h.
0.01
h
2
can never be equal to 0, the particle is never at rest.
f 1  h   f 1 
h
1
 sin 1.0001

 0.166
1
0.0001
0.0001
 0.172
1
 0.173
Spring 2014 BC 1 Semester Review Solutions p. 5
SP 14
12.
dy
dx
13.
1
2
  
 

1
  sec x  sin2 e x     sec x tan x  2sin e x  cos e x  e x 
2
y  sec x  sin2 e x  sec x  sin2 e x
1
2




f  x   ln x 2  3x  f 2  ln 22  3  2  ln 10 
f  x  
2
1
x  3x
 2x  3  f  2 
2
1
2  32
  2  2  3 
Therefore, the equation of the normal line is:
14a.
7
10
y  ln 10   
10
 x  2
7
14b.
15a.
f has a local maximum at x = 0.15 since f ′ changes from positive to negative at this point.
15b.
f is decreasing on the interval [0.15, 0.84] since f ′ is negative (nonpositive) on this
interval.
15c.
f is concave down on the intervals (–0.3, 0.5) and (1.4, 2) since f ′ is decreasing (and thus f
′′ is negative) on these intervals.
15d.
f has inflection points at x = –0.3, x = –0.5, x = 1.4, and x = 2, since f ′ has local maximums
or minimums (and thus f ′′ changes sign) at these point.
Spring 2014 BC 1 Semester Review Solutions p. 6
SP 14
15e
.

0.3
0.15
0.5
0.84
1.4
2
16.
Critical points are points where the derivative of a function is either zero or undefined.
17.
g  x   ln x 2  3x  4

g  x  
1
2
x  3x  4
2x  3
0
2
x  3x  4
2x  3  0
x 

For 0,3 ,
 2x  3
3
9 9

7
g    ln    4   ln  
2
4 2

4
g  0   ln  4 
g 3  ln  9  9  4   ln  4 
3
2
7
Global (absolute) maximum value of g: ln(4); Global (absolute) maximum value of g: ln  
4
18.
f  x   3sin x  2sin3 x
3cos x  0
f   x   3cos x  6sin2 x cos x

 3cos x 1  2sin2 x
x 

2
or x 
0
f
f
+
3
2


4
2
-

or x 
cos x  0

4
or x 
3
4
+
inc dec inc
dec
or
sin x  
3
5
7
or x 
or x 
on  0,2 
4
4
4
5
4
-
or 1  2sin2 x  0
3
2
+
7
4
-
1
2
2
+
inc dec inc
Spring 2014 BC 1 Semester Review Solutions p. 7
SP 14
5
7
.
,x 
2
4
4

3
3
Local maxima occur at x  , x 
.
,x 
4
4
2
Now, check values at these points an at endpoints to determine global maxs/mins.
 
 
 3 
 5 
 7 
 3 
f    1; f    f 
  2; f 
 f 
   2; f 
  1;f 0  f 2  0
2
4
 4 
 4 
 4 
 2 
Local minima occur at x 

,x 
 
 
5
7
.
and x 
4
4

3
2 occurs at x 
.
and x 
4
4
Global minimum value of  2 occurs at x 
Global maximum value of
19.
1
f  x   x 1  x  x 1  x 2
1
1

1
x
f   x   1  x 2  x  1  x  2   1   1  x 
2 1x
2

 f has a critical point at x  1 since f  is undefined at this value of x .
x
2
1x 
0
1
3
2 1x
2 1  x   x  0
2  3x  0
f
2
3
+
-
f
inc
dec
x 
Since f  is positive for all x 
2
2
and negative for all  x  1 ,
3
3
2
2 3
2 2 1
is the maximum value of f.
f 

or
9
3 3 3 3 3
20.
y  x 3  sin x  y   3x 2  cos x  3x 2  cos x  0  x  0.5354 or x  0.5354
21.
f  x   x 4  6x 3  7x  3
f   x   4x 3  18x 2  7
2
f   x   12x  36x
f   x   24x  36  f   0   36 and f  3  36
Since f  is negative at x  0, f  is concave down at
12x 2  36x  0
this point, meaning f  has a local max at x  0. Since
12x  x  3  0
f  is positive at x  3, f  is concave up at this point,
x  0 or x  2
meaning f  has a local min at x  3.
Spring 2014 BC 1 Semester Review Solutions p. 8
SP 14
22.
g  f 1  g   a  
 g  9 
23.
1

f f
a  
1
1

f  f 1  9 
 g  9 
1
f  5 
 g  9 
1
3
g  f 1  g   a  
 g  9 

1

f  f 1  a 

1
f f
1
Since f 5  9,f
1
9  5 

9
To find f 1  9  , first we need to solve 2x 5  3x  1  9 for x .
2x 5  3x  1  9  x  f 1  9   1.175


f   x   10x 4  3  f  f 1  9   22.05
 g   9   0.454
24.
9x 2  6 y  x 2y  17
9x 2  6 y  x 2y  17  18x  6 

dy
dy
 2xy  x 2 
dx
2 y dx
1

dy
dy
x2 
 2xy  18x
dx
y dx
3


dy  3

 x 2   2x  y  9 

dx  y


2x  y  9  y
dy 2x  y  9 


or
3
dx
3x2 y
x2
y

a)
2
2
9 1  6 y  1 y  17
Spring 2014 BC 1 Semester Review Solutions p. 9
SP 14
y 6 y 8  0

y 2


y 4 0
y  2 or
y 4
y  4 or y  16
2 1 16  9  16
dy

2
dx 1,16
3  1  16
 56
2 1   4  9  4
dy

2
dx 1,4 
3  1  4
 20
Equations of tangent lines are y  4  20  x  1 and y  16  56  x  1 .
b)
To have horizontal tangent line, must have 2x  y  9  0.
x  0 or y  9
2
2
9  0   6 y   0  y  17 or 9x 2  6 9  x 2  9   17
17
or 9x 2  18  9x 2  17
6
289
y 
or 18  17 no solution! 
36
 289 
Horizontal tangent line occurs only at  0,
.
 36 
y 
25a.
The function f(x) is not differentiable on any interval containing x = –1 or x = 2, so the
MVT cannot be applied here, meaning it can tell us nothing about f(x).
25b
Since the function f is continuous on a closed, bounded interval, the EVT guarantees that
f(x) has both a minimum and a maximum value of [-3, 5]. But since (-3, 5) is not a closed
interval, EVT guarantees nothing, and in fact the function does not have a maximum on
that interval.
26.
Average value of h on [-1, 3] is
h 3  h  1 
3   1 

47   5 
4
 13 .
h   x   3x 2  6
3c 2  6  13  c  
7
3
Only the positive square root is on [–1, 3], so answer is c 
Spring 2014 BC 1 Semester Review Solutions p. 10
7
or
3
21
.
3
SP 14
27.
The graph at right works for all three parts. It is not
continuous, so fails all the hypotheses. But it clearly hits
every y value between its endpoints (IVT), has an
instantaneous rate of change at least one point in the
interval that is equal to the average of change of the
function over the interval (MVT), and has a minimum and
maximum value (EVT),
28.
V 

3
r 2h and h  3r V   r 3 
We know
dV
dr
 3 r 2
dt
dt
dV
 12 and when h = 4, r = 4/3.
dt
2
 4  dr
dr
9
 12  3  


dt 4
 3  dt
 The radius is increasing at
29.
9
ft/sec.
4
In the diagram, the 300 is fixed, but the x and the y
dx
change. In fact, we are given that
is 8 ft/s.
dt
300
dy
dx
2
2
2
y  300  x  2y
 2x
dt
dt
dy
 2  500 
 2  400  8
dt
dy 32


dt
5
32
ft/sec .
Thus, the rate at which string is being let out is
5
Now, if  is the angle of elevation to the ground, we have:
300
d
300 dx
tan   
 sec2  

x
dt
x 2 dt
x
y
2
 500  d 
300


8

4002
 400  dt
d
6


dt 625
Thus, the angle of elevation is decreasing at
6
or about 0.096 rad/sec. .
625
Spring 2014 BC 1 Semester Review Solutions p. 11
SP 14
30.
V   r 2h 
dV
dr
dh
 2 r
h  r 2
dt
dt
dt
Since the volume of the cylinder is fixed,
2
dr
2     5   0.25 

dt
25
dr

 20
4
dt
dr
5


dt 16
5
ft/sec .
Thus, the radius is changing at
16
dV
 0.
dt
 0  2  5 
31.
Linearization, L, of g at x = 4:
L  x   g   4  x  4   g  4   L  x   3  x  4   6 or L  x   3x  18
L  4.2   3  4.2   18  5.4  g  4.2   5.4
Since g   x   5  0 for 2  x  6, g is concave up for 4  x  4.2.
Thus, the linearization of g at x  4, which is the tangent line to g at x  4,
will lie below the graph of y  g  x  , meaning that the estimate will be less than
the actual value of g  4.2  .
32.
1  f   x   4 and f 2   5  5  1  x  2   f  x   5  4  x  2 
 5  1  7  2  f 7   5  4 7  2
 10  f  7   25
33.
Suppose the printed part of the poster has dimensions x by y, with x going from left to
50
. . Since the side margins
right and y going from top to bottom. Thus, xy  50 or y 
x
are 2 inches, the width of the whole poster is going to be x + 4, and since the top and
bottom margins are 4 inches, the height will be y + 8. We need to minimize area.
 50

400
A  x 4 y 8 A  
 4 y  8  A 
 4y  32 .
y
 y





Differentiating, we find that A   

400
y2
 4.
Spring 2014 BC 1 Semester Review Solutions p. 12
SP 14
Setting this to zero yields y = 10, which gives a minimum because A 
800
y3
, which is
greater than 0 when y = 12. Then x is 50/10 = 5, the width is 5 + 4 = 9 inches, and the
height is 10 + 8 = 18 inches.
Spring 2014 BC 1 Semester Review Solutions p. 13
SP 14