Name
ANS WER KEY
Math 1205
Tes t 2
October 26 , 200 5
Class Time or CRN
READ THE DIRECTIONS . YOU MUST SHOW A L L WORK ON THIS TEST AND USE METHODS
LEARNED IN CLAS S TO RECEIVE FULL CREDIT. CALCULATORS ARE NOT PERMITTED.
( 100 pt.)
1 . Given the graph of the function below,
(5 pts)
Where is the function not differentiable? Explain why the function is not differentiable.
x= -2 and 4 – points of discontinuity,
x= -1 – corner,
x=2 – vertivccal tangent
x=5 - cusp
2 . Using the limit definition of the derivative, compute f !(x) if f (x) =
Show your work.
(
h!0
h!0
f '(x) =
(9 pt)
2 ( x + h ) + 1 " 2x + 1 # 2 ( x + h ) + 1 + 2x + 1 &
2 ( x + h ) + 1 " ( 2x + 1)
%
( = lim
h
$ 2 ( x + h ) + 1 + 2x + 1 ' h!0 h 2 ( x + h ) + 1 + 2x + 1
f '(x) = lim
f '(x) = lim
2x + 1 .
h
(
2x + 2h + 1 " 2x " 1
2 ( x + h ) + 1 + 2x + 1
)
= lim
2
2
=
=
2x + 1 + 2x + 1 2 2x + 1
h!0
h
(
1
2x + 1
2h
2 ( x + h ) + 1 + 2x + 1
)
= lim
h!0
(
)
2
2 ( x + h ) + 1 + 2x + 1
)
3 . Find the derivatives of each of the following functions using the derivative rules that we have had
up to this point. Simplify your answers.
(25 points)
5
f (x) = x10 + 5(x 2 ! 1) + e x ! 8 x + x 5 = x10 + 5x 2 ! 5 + e x ! 8 x + x 2
a)
3
f '(x) = 10x 9 + 10x + e x ! 8 x (ln 8) + 52 x 2
r(! ) = sec 2 (5! ) + tan 2 (5! ) = ( sec(5! )) + ( tan(5! ))
2
2
(
)
2
b) r '(! ) = 2 sec(5! ) ( sec(5! ) tan(5! )) (5) + 2 tan(5! ) sec (5! ) (5)
r '(! ) = 10 sec 2 (5! ) tan(5! ) + 10 sec 2 (5! ) tan(5! )
r '(! ) = 20 sec 2 (5! ) tan(5! )
y=
x + cos(x)
1 + sin(x)
(Use Quotient Rule)
c) y' = (1 + sin(x)) (1 ! sin(x)) ! ( x + cos(x)) ( cos(x))
2
(1 + sin(x))
y' =
(
1 ! x cos(x) ! sin 2 (x) + cos 2 (x)
(1 + sin(x))
)
2
(1 + sin(x))
2
( )
+ 1) (10x + x
f '(x) = e x !" 4(x 5 + 1)3 5x 4 #$ + (x 5 + 1)4 !" e x (2x) #$
2
2
f '(x) = 2xe x (x 5
y = 2 sin !1
e) y' = 2
y' =
3
3
2
5
)
+1
( x)
1
1!
( x)
1
x ! x2
2
" 1 %
$#
'
2 x&
=
1
x 1! x
1 ! sin 2 (x) ! x cos(x) ! cos 2 (x)
(1 + sin(x))2
1 ! x cos(x) ! (1)
=
f (x) = e x (x 5 + 1)4
d)
=
2
=
!x cos(x)
(1 + sin(x))2
4 . Find the first and second derivatives of w = tan !1 (x 2 ) .
w '(x) =
(10pts)
1
2x
( 2x ) =
2 2
1 + (x )
1 + x4
(1 + x )( 2 ) ! 2x ( 4x )
w"(x) =
(1 + x )
4
3
=
4 2
2 + 2x 4 ! 8x 4
(1 + x )
4 2
=
2 ! 6x 4
(1 + x )
4 2
=
2(1 ! 3x 4 )
(1 + x )
4 2
5. Let f and g be differentiable functions with values in the table below.
g(x)
x
f (x)
f !(x)
g!(x)
1
6
2
3
1
2
4
1
8
2
(8 pt.)
a) Find h '(1) , where h(x) = f (x)g(x) .
h '(x) = f (x)g '(x) + g(x) f '(x)
h '(1) = f (1)g '(1) + g(1) f '(1)
h '(1) = (6)(1) + (2)(3) = 6 + 6] = 12
b) Find k '(2) , where k(x) =
k '(x) =
k '(2) =
k '(2) =
f (x)
g(x)
g(x) f '(x) ! f (x)g '(x)
( g(x))2
g(2) f '(2) ! f (2)g '(2)
( g(2))
2
(1)(8) ! (4)(2)
(1)
2
=
=
8!8
1
g(2) f '(2) ! f (2)g '(2)
( g(2))2
= 0
6 . For what values of x in [0,2! ] does the graph of f (x) = x + 2sin x have a horizontal tangent?
Include the quadrant(s), ! (the reference angle), and the values for x that answer the question.
(10pts)
f (x) = x + 2 sin(x)
f '(x) = 1 + 2 cos(x) = 0
2 cos(x) = !1
!1
cos(x) =
2
quadrant(s): II & III ,
! (the reference angle):
!
,
3
values for x=
2!
4!
&
3
3
7 . The equation xe y = y ! 1 defines y as an implicit differentiable function of x.
a. Using implicit differentiation, find
(10 pts)
dy
in terms of y and x.
dx
xe y = y ! 1
x[e y (y')] + e y [1] = y'
e y x(y') ! y' = !e y
y'(e y x ! 1) = !e y
!e y
y' = y
(e x ! 1)
b. Write the equation of the tangent line at (0,1).
y'(0,1) =
!e1
!e
=
=e
1
(e (0) ! 1) !1
tangent line : y ! 1 = e(x ! 0) " y = ex + 1
8. The graphs of two functions are given below. Decide if each function has an inverse function. If
the inverse function exists, state the approximate domain and range of the inverse function. If no
inverse exists, leave the domain and range blank.
(6pts)
Does f(x) have an inverse?
No
Does f(x) have an inverse?
Range of inverse
Range of inverse:
Domain of inverse
Domain of inverse:
[0 , 1.4]
[1 , 3]
Yes
9 . The position of a particle for 0sec. ! t ! 3sec. is s(t) = 2t 3 ! 6t 2 ! 2 feet.
(12 pts)
v(t) = 6t 2 ! 12t = 6t(t ! 2) = 0 " t = 0 & 2sec
v _____! ______ | ______+ ______
0
2
3
a. When is the particle at rest?
t=0 & 2 sec
b. What is the position when the particle is at rest?
c. When does the particle move to the left?
s(0)=-2 & s(3)=-2
(0, 2)
d. Draw a horizontal line graph of the particle’s position and direction for 0sec. ! t ! 3sec. .
s(3)
" !! > ! ! ! ! ! > ! ! !•
$# • ! ! < ! ! ! ! ! < ! !•
-10
s(2)
-2
s(0)
e. Find the total distance traveled.
Total distance = distance left + distance right
= |-2-(-10)| + |-10-(-2)| = 8 + 8 = 16 ft
10 . The following graph is the graph of y = f (x) .
Carefully draw the graph of the derivative, f '(x) on the same axes.
(5 pts)
Bonus: The figure shows the graphs of f , f !, and f !! . Identify each curve.
a =
f”
b=
f
c=
f
Pledge : I have neither given nor received help on this tes t .
Signed
(3 pts)