AP Calculus Holiday Packet Solutions:
h "( x ) =
f ( x ) = 1+ x "2
f #( x ) = "2x "3
1: c
!
6
f ##( x ) = 4
x
f ##( x ) > 0 if x $ 0
h "( 3) =
2: b
#2!
3
f "( x ) = (9 # x 2 ) 5 (#2x )
5
#6x
f "( x ) =
2
5(9 # x 2 ) 5
4: d C.V .:x = #3,0,3
!
16 (10 " 2) " (10 " 8)
=
!
x2
6
16
=1
x2
7: a x = ±4, only 4 in interval
h "( 3) =
f "( x ) =
g( x ) f "( x ) # f ( x ) g"( x )
[g( x )]
f "( x ) =
2
f "( x ) =
4
( x + 3)
2
= 4( x + 3)
!
f "( x ) = #3x 2 + 36x #105
f "( x ) = #3( x 2 #12x + 35)
f "( x ) = #3( x # 5)( x # 7)
"
10:d f ( x ) > 0 on (5,7)
!
2
#x 2 # 2
[( x + 2)( x #1)]
2
!
# dy
&
dy
10x " 2% x
+ y ( + 14 y
=0
$ dx
'
dx
dy
10x " 2y = (2x "14 y )
dx
dy 5x " y y " 5x
=
=
dx
x
"
7y
7y " x
6: b
!
y " = 1+ cos x = 0
x =#
By sign chart, 1+ cos x > 0
9: e for all values of x $ #
3
Denominator is negative if
5: d x > #3
f ( x ) = "6x "1
f #( x ) = 6x "2
"12
x3
if x < 0, f ##( x ) > 0
f ##( x ) =
8: a
x3
f "( x ) =
# 4x
3
f ""( x ) = x 2 # 4
y " = 1+ sin x = 0
!
+ x # 2]
#2
#8
( x + 3)
2
Numerator is always negative and
denominator is always positive
3: d except for x = #2,1
1(6) # 0( 13 )
=6
12
f "( x ) =
[x
2
g( 3) f "( 3) # f ( 3) g"( 3)
[g(3)]
x 2 + x # 2 # x (2x + 1)
!
3#
2
By sign chart, 1+ sin x > 0
3#
for all values of x $
2
11:e
x=
f "( x ) = 15x 4 # 15x 2
!
2
f "( x ) = 15x ( x + 1)(x # 1)
concavity change at x = ±2
16
25
f (±2) = # 8 + 15 =
12
3
12:d
!
v ( t ) = 6t " 3t 2
v #( t ) = 6 " 6t = 0
13: c
!
f"
14: b
++++ -------- ------- ++++
-1
0
1
t =1
15:c
v (1) = 3
!
!
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Stu Schwartz
f ( x) = x 2 " 4 x + 7
!
16:c
Synthetic, f ( x ) = "x 2 + 2x + 1
2
f "( x ) = ( x + 1) + 2( x + 1)( x # 2)
f #( x ) = 2x " 4 = 0
f !( x ) = "2x + 2 = 0
CV:x = 2
CV:x = 1
f "( x ) = ( x + 1)[( x + 1) + 2( x # 2)]
f # ---------- ++++++
0
2
3
#
f (0)DNE, f (2) = 3, f (3) = 4
f +++++++++ -------
f "( x ) = ( x + 1)( 3x # 3)
!
1
-1
17: e
f "( x ) = 3( x + 1)( x #1)
2
f ("1) DNE, f (1) = 2, f (1)
"
18: a Create sign chart for f ( x )
!
Synthetic, f ( x ) = x 2 + 4x + 3
!
3
f "( x ) = 4 x #18x
!
f !( x ) = 2x + 4 = 0
2
CV:x = #2
2
f ""( x ) = 12x # 36x
f --------------- ++++
!
f ""( x ) = 12x ( x # 3)
19:d
"
f (0) = 0, f ( 3) = #81
-2
-4
-1
f (#4 )DNE, f (#2) = #1, f (#1) = 0
20: a
dy
= sec 3x tan 3x ( 3)
dx
21:b dy = 3sec 3x tan 3xdx
f "( x ) = 4 x 3 + 3x 2
"
!
f ""( x ) = 12x 2 + 6x
f ( x ) is a parabola
22: c so f is decreasing on (6,")
!
5
3
5
2
2c 2 = 3 # c = ±
=
f ( x ) is a parabola
f ""( x ) = 6x (2x + 1)
opening downward
2 " 13
1
=
=
c 2 " 12 + 3
!
2
3
opening downward
$ 1'
1
f (0) = 0, f & # ) = #
% 2(
16
23:a
24: b so f is decreasing on (9,")
!
!
Work backwards
Possible 1st derivative is
3
2
$ "1'
3
On &"3, ),c = "
%
2(
2
25:a
f "( x ) = x 4 # x 2 = x 2 ( x + 1)( x #1)
f "(#3) = 0
27: a Refer to sign chart on #14
26:c
f "( x ) = 3x 2 + 1 = 0
!
x =±
!
!
1
, via sign chart
3
Since the endpoints are
open, there is no minimum
The graph of f is concave
1
1
rel max:x =
,rel min:x = #
3
3 29: b value. There is a max at x = c. 30:b up at x = "1, so f ##("1) > 0
28:e
!
!
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!
Stu Schwartz
f "( x ) = 2( x + 1)( x # 3) + ( x # 3)
2
f "( x ) = ( x # 3)[2( x + 1) + x # 3]
f "( x ) = 3( x + 2)
If f ""is a positive constant
x = #2 is a CV but
then f " is a positive linear
"
f ( x ) > 0 if x $ 2 so
so f must be parabolic
32:c function always increasing 33:a opening up.
f "( x ) = ( x # 3)( 3x #1)
f ""( x ) = 6x #10
31:b No inflection pt at ( 3,0)
If f "(#2) = 0, then there is
!
2
If f "( 4 ) = 2, then the
a horizontal tangent at x = #2.
!
function is increasing!at
Since f ""(#2) > 0, then f is concave
x = 4. Since f ""switches signs
34:b up at x = #2, meaning a rel min.
35:d at x = 4, it is an inflection pt. 36:c
!
!
Since there are asymptotes
f (0) = 0, f (2) = 10
37:d f (0) " f (2)
at x = ±2, the answer must
38:c be "inside" the asymptotes. 39: c
!
!
40:d
This curve has a horizontal asymptote
at y = 1 (powers equal). The larger the
value of a, the slower it takes for the
curve to approach the aymptote.
!
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Stu Schwartz