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```CALCULUS I with TOPICS TRIMESTER 3 REVIEW SOLUTIONS
1.
a.

June 2008

( x  x)3  5( x  x)  ( x 3  5 x)
f ( x  x)  f ( x)
 lim
x 0
x 0
x
x
3
2
2
3
3
x  3x x  3xx  x  5 x  5x  x  5 x
 lim
x 0
x
2
2
3
3 x x  3xx  x  5x
 lim
 lim 3 x 2  3 xx  x 2  5
x 0
x 0
x
2
 3x  5
f ( x)  lim
b.


( x  x)  2  x  2
f ( x  x)  f ( x)
 lim
x 0
x  0
x
x
 ( x  x)  2  x  2  ( x  x)  2  x  2 


 lim 
x 0 
 ( x  x)  2  x  2 
x



 x  x  2   ( x  2)
x
 lim
 lim
x 0
x  0
x ( x  x)  2  x x  2
x
( x  x)  2  x  2
f '( x)  lim



 lim
x 0
2.

1





( x  x)  2  x  2
a.
f  x 





2

1
x2



 x  2  , discontinuous at x 5, x 2 ; not differentiable at
 x  2  x  5
x 5, x 
2
b.
3.
4.
continuous everywhere, not differentiable at
x 0 .
a.


b.
f   x   3x 2  1 e x
a.
y 
8 x 3 4 x
b.
y  2k sin(kx) cos(kx)  3 tan 2 x sec 2 x
3
x
CALCULUS 1 TRIMESTER 3 REVIEW SOLUTIONS June 2008
page 1
4. continued
c.
e3 x
, x0
y  3e ln(7 x) 
x
d.
y 
e.
y  13 tan 6 x 2  x
f.
y 
g.
5.
3x
2
 x  1
2
,
x  1 if simplified completely.
  

12
1 

2
2
12 x sec 6 x 

2 x

 
2 x cos x  x 2 sin x
2 x 2 cos x
a ln( x  c )
ax  b

xc
y   2 ax  b
 ln( x  c) 2
a.
y  9 x 2 ; Tangent line is y  36 x  54
b.
  1

y  4 cos  4 x  ; Tangent line is y  2 3  x   
24  2

6.
P  t  
P  t  
200t
t
2

1
2
600t 2  200
t
2

1
3
Five seconds after the blood leaves the heart, P(t) is changing at a rate of
P  5  1.48 , so blood pressure is decreasing at
a rate of -1.48.
P  t  is negative for 0  t  10, so P  t  is decreasing over this interval. P(t) has a horizontal tangent line at t=0, then
P(t) begins to decrease. From
P  t  , we see that the function has an inflection point where t  1 3 , which is where the
function decreases most rapidly. To the right of t  1
level off to its asymptote at
7.
3 , the function begins to decrease more gradually as it begins to
P  t   25 .
a. g ( x)  3 x  6 x; g ( x)  6 x  6
2
g is increasing when x  2 and when x  0 , g is decreasing when 2  x  0 , g is concave up when x  1 and
concave down when x  1. So, g has a local min at x  0 , a local max at x  2 , and a point of inflection at x  1 .
Graph the function on your calculator to check your sketch of the graph.
CALCULUS 1 TRIMESTER 3 REVIEW SOLUTIONS June 2008
page 2
7. continued
b.
dy x 2  2 px dy
;
changes signs when x  0 or x  2 p .

dx ( x  p)2 dx
d2y
2 p2

dx 2  x  p 3
d2y
;
changes sign when x   p
dx 2
x  0 and decreasing
when 2 p  x  0 . The function is concave up when x   p and concave
down when x   p . So, the function has a local min at x  0 and a local max
at x  2 p . There is a vertical asymptote at x   p .
The function is increasing when x  2 p and when
8.
The rate of change in gasoline prices is positive but less positive than it was previously. In calculus terms:
increasing) and
9.
x  p
G  t   0 (prices
G  t   0 (concave down; increasing at a slower rate)
a. When the outside temperature is 23 degrees Celsius, the cost of heating my house is \$7.54 per day.
b.When the outside temperature is 23 degrees Celsius, then the cost of heating my house is decreasing by \$0.17 a day per
degree Celsius.
c. The line tangent to f at
Then
So
T  23 is y  0.17 T  23  7.54 .
f  21  0.17  21  23  7.54 .
f  21  7.88 dollars per day.
f  2   f 1  ln 2    ln1
2
10. a. Average rate of change is

  ln 2 
2 1
1
2
1
b. Based on the table below, the instantaneous rate of change of f at x = 2 is approx. 0.517.
f  2  h   f  2
h
 ln  2  h  

2 h
  ln 2 
2
h
f  2  h   f  2
h
h
0.01
0.001
0.0001
-0.0001
-0.001
-0.01
0.5191
0.5172
0.5171
0.5170
0.5169
0.5151
CALCULUS 1 TRIMESTER 3 REVIEW SOLUTIONS June 2008
page 3
10. continued
c. Let y   ln x  .
x
ln y  ln  ln x   x ln  ln x 
x
1 dy
1 1
1
 ln  ln x   x
 ln  ln x  
y dx
ln x x
ln x
dy
1 
1 
x

 y  ln  ln x  
   ln x   ln  ln x  

dx
ln x 
ln x 


dy
1 
2
  ln 2   ln  ln 2  

dx x  2
ln 2 

11. If he rows the whole way, he will row 680 ft. If he rows directly to shore, and then walks to the cabin, then he will row 330 ft and
walk
6802  3302  594.56 ft.
x = the distance from the cabin that he puts to shore. (Domain of reality is 0  x  594.56 .)
x
Then the distance Bill walks is Dwalk  x , and so the time he spends walking is Twalk 
seconds.
2.8
Let
The distance he rows is Drow 
(594.56  x) 2  3302 , and so Trow 
Then the model for his total travel time as a function of
x is T ( x) 
(594.56  x) 2  3302
.
2.2
(594.56  x) 2  3302
x
.

2.8
2.2
We want to find both the absolute minimum and absolute maximum travel times, and the values of
x which will produce them.
We need to consider how much time the trip would take at either of the endpoints of the domain of reality, because absolute
extrema may occur at endpoints of domains!
x  0 then he is rowing the entire way, and T (0)  309 seconds.
If x  594.56 then he is rowing directly to shore and walking the entire shore, and T (594.56)  362 seconds.
If
But we could have local min’s and max’s above 362 secs or below 309 secs, so now we must find any relative extrema …
1
(594.56  x)

. Find all x -values for which T ( x)  0 or T ( x)
2.8 2.2 (594.56  x) 2  330 2
DNE. The only critical number is x  175.4 ft.
Differentiate w.r.t.
x to find T ( x) 
We must use the first or second derivative test to determine whether this value produces a relative max, min, or neither.
The derivative, T ( x)  0 for
x  175.4 , and T ( x)  0 for x  175.4 . Hence we know that for x  175.4 ft, Bill’s total
travel time will be minimized. It would take T (175.4)  305 seconds (i.e. 5 minutes 5 seconds).
So, we can conclude that in order to minimize his traveling time, Bill should row directly to the point on shore that is 175.4 ft from
his cabin, and that this should take 5 minutes and 5 seconds. To maximize his travel time, he should row directly to shore and
walk the entire shoreline, and that this path would take 6 minutes and 2 seconds.
CALCULUS 1 TRIMESTER 3 REVIEW SOLUTIONS June 2008
page 4
x  2x  3

2
12. a. lim
has
x 1
x 
3 x  22 x  100

2
b. lim
5x  7 x  1
x 
2
10  3e
c. lim
e 4
x 
tan( )
d. lim
 0
x
4
1  cos(3 x )


x  2x  3
 0
 lim
7
6
x  10
sec
 lim
 0

2

1
1
  so limit does not exist
 
2
t 
x  ln x
tan 
lim
t
3 x ln x
6 x  22
 lim
x  10 x
2
2
2
x 1

1
5
3
  but lim
This has indeterminate form
x 
5x  7 x  1
x 
t 
h. lim
x 
2
lim (t  2) e has indeterminate form 0   so rewrite: lim (t  2) e
2
2x  2
 lim
3 x  22 x  100
indeterminate form so apply l'hopital's rule: lim
indeterminate form so apply l'hopital's rule:
0
x  2x  3
x
x 0
0
3
e e

04
2
x 1
g.
has

e. lim
f. lim

x 1
x 
10  3(0)
x
x
has
x  2x  3
2
indeterminate form so apply l'hopital's rule: lim

has



t
 lim
(t  2)
t 
e
t
t 
t 
indeterminate form so apply l'hopital's rule:
lim
x 
t
(t  2)
so apply l’hopital’s rule: lim (t  2) e  lim
2
2
e
3 x ln x
x  ln x
2 t  2
2
 lim
t 
t
 lim
e
3 ln x  3
x 
1
1
 lim
t 
t
2
e
t
0

x
1
2
i. lim
4 y
y 0
has
u  u 1
2  2 1
2

u  2u
2
9 3
x
x 0
x
has
x
9 3
x
lim
x 0

x  
l. lim  1 
2  2(2)
2
0
2
4 y
 lim
y 0
y

1
4
1
8
x
9 ln 9  3 ln 3
x
 lim
x
x 0
 ln 9  ln 3  ln(9 / 3)  ln 3
1
x
0

 has 1 indeterminate form so use logs, then rewrite to get indeterminate form & apply l'hopital's rule:
x
0
  
lim 1 
 x 

7

1
ln 
2 4 y
indeterminate form so apply l'hopital's rule:
0
x
2
y 0
0
u 2
k. lim
indeterminate form so apply l'hopital's rule: lim
y
2
j. lim
0
2
x
x

lim
x 


ln  1 
2
x
 2
x  ln  1    lim
  xlim

x
 x  x


ln  1 
1
x

x  
So, lim  1 
2
2

x

1
lim
x 
2

x
1
x
2
2
x2

lim
x 
2
1
2
 2
x
x
2
 e
x
CALCULUS 1 TRIMESTER 3 REVIEW SOLUTIONS June 2008
page 5

3
5
13. a.
1
3
14. a.
f ( x)  5 x
b. 
2
1
c. 0
3
d. 0
5
 x 3 ; f ( x) 
3
e. No
10  5 x
3x
changes sign at
1
; f ( x ) changes sign at
4
5
x  2 and at x  0 ; f ( x) 
x  1 . The domain of f is all real numbers . f is increasing when

10  x  1
3
as f (0) is undefined. f is concave up when
b. f ( x)  x
f. No
0 x2
9x
4
; f ( x)
3
. f has a “cusp” at (0,0)
x  1.
4
4 1 4
4
x ; f ( x)  x 5  ; f ( x ) changes sign at x  0 and at x  1 .; f ( x)  
;
6
5
5
5
5
25 x
f (0) undefined and f ( x)  0 for all non-zero real numbers. The domain of f is all real numbers. f is increasing when
x  1 and when x  0 . f has a “cusp” at (0,0) as f (0) is undefined. f is never concave up.
15.
2 xy  x 2 y   2 yyx  y 2  0
y  x 2  2 xy    y 2  2 xy
 y 2  2 xy
x 2  2 xy
1  4 3
y  2,1 

 undefined
44 0
y 
So, the tangent line is vertical:
x  2.
16.
5000
 5000

C ( x)  0.25  3 x 2  0.1  2  3 x  2 
 x  2
2
3x
 3x

4000
 0.25  3 x 2 
3x
Dimensions:
4000
C '( x)  0.75  2 x 
Length: 28.845 cm,
3x 2
Width: 9.615 cm
C ( x )  0  4.5 x 3  4000  0
Height: 18.028 cm
x  9.615cm
8000
C ( x )  15 x 
, second derivative is positive for all positive x-values so at x = 9.615 C(x) has a minimum.
3x 3
17. Graph of f :
CALCULUS 1 TRIMESTER 3 REVIEW SOLUTIONS June 2008
page 6
18. a. f ( g (4))  f (3)  4
b. h(2)  f ( g (2)) g (2)  f (1)  2  2
g (4) f (4)  f (4) g (4) 3  3 1 1 8


( g (4)) 2
9
9
d. h '(3)  g '( f (3)) f '(3)  g '(4) f '(3)  2
e. h(3)  f (3) g (3)  f (3) g (3)  2  4  4  3  20
c.
f.
19.
h(4) 


1
 1

h(1)  h(1)  g (1) 
 f (1)  g (1)  ln  f (1)    f (1) g (1)  2  1  4  ln 3   6  36ln 3
f (1)
 3



 a
a
  bx  1 for x  1
  b for x  1
so
f ( x)   x
f ( x)   x2
2ax 2  3b for x  1
4ax for x  1

a  b  1  2a  3b
a  b  4a
1  3a  4b
a
0  3a  b
20.
1
9
b
1
3
f   x   aebx  abxebx  aebx 1  bx 
aebx 1  bx   0
1  bx  0  x  
Max at x 
1
b
1
1
   b  3
3
b
1
 1  3  a
Max value : 1  a   e  3   e 1  a  3e
3
3
CALCULUS 1 TRIMESTER 3 REVIEW SOLUTIONS June 2008
page 7
```
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