Ap® CALCULUS AS
2004 SCORING GUIDELINES
dy
(a) Show that -d
x
=
3y - 2x
8
".
y -.>x
(b) Show that there is a point P with x-coordinate
horizontal. Find the y-coordinate of P.
3 at which the line tangent to the curve at P is
2
(c) Find the value of d ; at the point P found in part (b). Does the curve have a local maximum, a
dx
local minimum, or neither at the point P ? Justify your answer.
(a)
2x
+ 8yy'
+ 3xy'
= 3y
(8y - 3x)y'
2 : { I : implicit djff~rentiation
I : solves for y
= 3y - 2x
3y - 2x
y'=--8y - 3x
3y - 2x
--= o· 3y - 2x = 0
8y - 3x
'
=
When x
3, 3y
=
1:dY=O
dx
.>:
1: shows slope is 0 at (3, 2)
{
1 : shows (3, 2) lies on curve
"
6
y=2
Therefore,
P = (3,2)
is on the curve and the slope
is 0 at this point.
(c)
d2y
dx2
At P
- 2) - (3y - 2x)(8y'
(8y - 3x)(3y'
- 3)
(8y-3x)2
= (3,2),
2
dy
dx2
=
(16 - 9)( -2)
(16-9)2
= _~.
7
Since y' = 0 and y" < 0 at P, the curve has a local
2'
2
dy
. dx2
I : value of
---+
d2
at (3,2)
dx
1 : conclusion with justification
maximum at P.
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Ap® CALCULUS AS
2005 SCORING GUIDELINES
v(t)
A car is traveling on a straight road. For 0 :-::::
t :-::::
24 seconds, the car's
:g 20
vet), in meters per second, is modeled by the piecewise-linear
velocity
function defined by the graph above.
;;.....g
.::: V>
~ vet) dt. Using correct units, explain the meaning of Jo
~ vet) dt.
(a) Find Jo
(b) For each of v'(
4) and v'(20),
15
~8.1O
"'",
>b
~
find the value or explain why it does not
g
5
4
exist. Indicate units of measure.
8 12 16 20 24
Time (seconds)
at time t, in meters per second per second. For 0 < t < 24, write a
(c) Let aCt) be the car's acceleration
piecewise-defined
(4,20)
o
function for a( t).
(d) Find the average rate of change of v over the interval 8:-:::: t :-::::
20. Does the Mean Value Theorem guarantee
c, for 8 < c < 20, such that v'( c) is equal to this average rate of change? Why or why not?
a value of
(a)
S:4 v(t)
dt
=
t( 4)(20) + (12)(20) + t(8)(20)
=
360
The car travels 360 meters in these 24 seconds.
(b)
v'( 4) does not exist because
lim (V(t)
1-74-
V
'(
20
)
aCt)
=
- V(4))
t- 4
=
{
-'2~
I:
5;f:. 0
=
lim (V(t)
1-74+
20 - 0
= 16-24
=-'25
5
/
m sec
3:
- V(4)).
t- 4
2 :
= 4 and
t
-
2.
/
-'25
1 : identifies constants with correct intervals
= 16.
I
The average rate of change of v on [8, 20] is
20 - 8
1 : units
1: finds the values 5, 0,
{
if 4 < t < 16
if 16 < t < 24
v(20) - v(8) __
{
v'( 4) does not exist, with explanation
I: v'(20)
2
if 0 < t < 4
aCt) does not exist at t
(d)
I: value
2' {
. 1: meaning with units
: average rate of change ofvon
2: {
I : answer with explanation
2
[8,20]
6 m sec .
No, the Mean Value Theorem does not apply to v on
[8,20] because v is not differentiable at t = 16.
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6
(for AP students and parents).
Ap® CALCULUS AS
2005 SCORING GUIDELINES
x
f(x)
f'(x)
f"(x)
O<x<1
I
-I
Negative
0
Positive
2
4
Positive
0
Positive
-2
Negative
0
Positive
0
1<x<2
3
3<x<4
Positive
0
Negative
DNE
Negative
-.)
Negative
DNE
Negative
0
Positive
2
2<x<3
"
Let f be a function that is continuous on the interval [0, 4). The function f is twice differentiable except at x = 2. The
function f and its derivatives have the properties indicated in the table above, where DNE indicates that the derivatives of f
do not exist at x = 2.
(a) For 0 < x < 4, find all values of x at which f has a relative extremum. Determine whether f has a relative maximum
or a relative minimum at each of these values. Justify your answer.
(b) On the axes provided, sketch the graph of a function that has all the characteristics of f
(Note: Use the axes provided in the pink test booklet.)
(c) Let g be the function defined by g(x) =
o<x
r
f (t) dt
on the open interval (0, 4). For
< 4, find all values of x at which g has a relative extremum. Determine whether g has a
relative maximum or a relative minimum at each of these values. Justify your answer.
(d) For the function g defined in part (c), find all values of x, for 0 < x < 4, at which the graph of g has a point of
inflection. Justify your answer.
(a)
f
has a relative maximum at x
=
2 because
I'
relative extremum at x = 2
I:
changes from
2: {
positive to negative at x = 2.
I : relative maximum with justification
points at x = 0, 1, 2, 3
I:
2:
(c)
g'(x) = f(x)
{ I : appropriate increasing/decreasing
and concavity behavior
I: g'(x)
= 0 at x = 1,3.
3:
g' changes from negative to positive at x = 1 so g has a relative
minimum at x
and behavior at (2,2)
=
1. g' changes from positive to negative at x
=
{
3
=
I(x)
1: critical points
I : answer with justification
so g has a relative maximum at x = 3.
(d)
The graph of g has a point of inflection at x
=
2 because g"
changes sign at x = 2.
= I'
I
:x =
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5
2
2: { I: answer with justi fication
(for AP students and parents).
Ap® CALCULUS AB
2001 SCORING GUIDELINES
Let h be a function
x2
2
by h'(x) = --x
(a)
defined for all x += 0 such that
for all x +=
h( 4) = -3
and the derivative
o.
Find all values of x for which the graph of h has a horizontal
whether
Justify
of h is given
h has a local maximum,
a local minimum,
or neither
tangent,
and determine
at each of these values.
your answers.
(b)
On what intervals,
if any, is the graph of h concave up? Justify your answer.
(c)
'Write an equation
for the line tangent
(d)
Does the line tangent
to the graph of h at x
to the graph of h at x
=
=
4.
4 lie above or below the graph of h for
x> 4? Why?
1 : analysis
0
h'(x)
I
x
+
--12
und
I
0
at x=--!2
Local minima
-
2
I
-!2
+
and at x
h"(x) = 1 + 2" > 0 for all x
x
0
2 : conclusions
< -1 > not dealing with
discontinuity
at 0
= -!2
the graph of h is concave up for all x +=
h'(4)
(d)
o.
3
1
1 :
h"(x) > 0
= 16 - 2 = !...
4
y
1 : h"(x)
+= O. Therefore,
7
+ 3 = -(x
2
The tangent
2
- 4)
line is below the graph because
the graph of h is concave up for x
>4.
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