APCalculus
APPracticeTest# Part I: Multiple Choice (no calculator)
1) If y = x sin x, then dy/dx =
A) sin x + cos x
B) sin x + xcos x
C) sin x – xcos x
D) x(sin x + cos x)
E) x(sin x – cos x)
2) If f (x) = 7x – 3 + ln x, then
∙
∙
AnswerB
1
A) 4
B) 5
C) 6
D) 7
E) 8
1
–
1
AnswerE 3) The graph of function f is shown. Which of the following statements is false?
(E) The function f is continuous at x = 3. Comments:
(A) TRUE. The limit exists because the function approaches the same value from the left and right. Note that . So, as an the value of the function is different than the value of the limit →
aside, we note that the function is not continuous at 2. (B) TRUE. The limit exists because the function approaches the same value from the left and right at 3. (C) FALSE. The limit does not exist because the function approaches different values from the left and right at 4. (D) TRUE. The limit exists because the function approaches the same value from the left and right at 5. (E) TRUE. The function is continuous because a) the function approaches the same value from the left and right at 3 and b) that limit is the value of the function at 3. AnswerC
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APCalculus
APPracticeTest# 4) If y = (x3 – cos x)5, then y’ =
for
5)
for
–
′
– ∙
– ∙
– AnswerE
2
2
Let f be the function defined above. For what value of k is f continuous at x = 2?
A) 0
B) 1
at all values of
C) 2
D) 3
E) 5
To get the value of at 2 for which the function is continuous, calculate the limit as → 2. lim
→
lim
1
lim
2
1 →
2
2
→
2 ∙ 2
2
1
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AnswerE
APCalculus
6) If
APPracticeTest# √
4and
3
2,then the derivative of f (g (x)) at x = 3 is
Use the chain rule: 1
∙
2
: First, find Then, find 4
3
2
3
2
4
4
∙ 2
2 3
∙
∙3
∙ 3 ∙
√
7)
Notice that the expression is the definition of a derivative:
ln 4
ln 4
→
1
4
ln
4
∙
3 into the chain rule expression for: Finally, substitute lim
4
√
: 3
∙
AnswerB
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∙3
√
∙ 3 √
AnswerA
APCalculus
APPracticeTest# 8) The function f is defined by
. What points (x, y) on the graph of f have the property that the line tangent
to f at (x, y) has slope of ?
Recall that the slope of the tangent line of a function at a point is equal to the derivative of the function at that point. Then, 2 ∙ 1 ∙ 1
2
2
2
2
2
2
The problem tells us that this is equal to a slope of . So, 2
1
2
2
4
4
0
4 4 0, 4 To get the points, substitute the ‐values into to get the points: ,
,
,
AnswerC
9) The line y = 5 is a horizontal asymptote to the graph of which of the following functions?
Horizontal asymptotes occur at the limits as → ∞ and as → ∞.
Let’s try limits as → ∞. (A) lim
→
(B) lim
→
(C) lim
→
(D) lim
→
(E) lim
→
0 5
∞ 0 lim
→
5 using L’Hospital’s Rule using L’Hospital’s Rule AnswerE
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APCalculus
APPracticeTest# 10)
If f ( x )  16 x then f "(4) is equal to
A
(B) -32
1
2
16 ∙
4
4
2
11) If
2
2
2
4
4
8
∙
(E) -16
8
1
2
8 ∙
(D) -2
16
4
(C) -4
4
AnswerA , whatisthevalueof
atthepoint 3, 0 ?
∙
Notice that, at the point (3, 0), 2
∙
∙
2
2
∙
2
2
2
∙ 2
2
∙ 1
2
2
,
Now, substitute in
2∙
,
∙ 2
2∙
2∙
∙ 1
2∙
3∙0
6∙5
9
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AnswerA APCalculus
APPracticeTest# 12)
d sin x
e  
dx 
A (C) cos
(B)
∙
∙ cos 13)
If y 
(A) 
sin
2
3
(B) 
1
3
sin AnswerD
7
(3x  1) 2
3
1 ∙
2
1 ∙
2
1
6x  5
(3 x  1) 2
(D)
∙
∙ 3
1
3
1
7
(3 x  1) 2
(E)
7  6x
(3 x  1) 2
1
2
3
(C)
3
(E)
2 x
dy
, then
=
3x  1
dx
9
(3x  1) 2
cos
(D)
3
1
6
1
3
3
AnswerB Part II: Graphing Calculator is Allowed
14) Let f be a function that is continuous on the closed interval [2, 4] with f (2) = 10 and f (4) = 20. Which of the
following is guaranteed by the Intermediate Value Theorem?
A) f (x) = 13 has at least one solution in the open interval (2, 4). TRUE since 13 is between 10 and 20. One way to think about it is that to get from 10 to 20, the function must go through 13. Answer A
B) f (3) = 15 FALSE. 3 can be anything, including values below 10 or above 20.
C) f attains a maximum on the open interval (2, 4). FALSE. may have a maximum at 4. For example, consider the line connecting (2, 10) and (4, 20). It has a maximum at 4, not in the open interval (2, 4).
D) f ‘(x) = 5 has at least one solution in the open interval (2, 4). TRUE, but this is not required by the Intermediate Value Theorem, which deals with the values of the function, not the values of the derivative.
E) f ‘(x) > 0 for all x in the open interval (2, 4). FALSE. The curve could slope downward in some sub‐interval between 2 and 4.
Intermediate Value Theorem: If a function, , is continuous on the interval ,
and , then there is a value in , such that value between Page 6 of 11
, and is a
. APCalculus
APPracticeTest# 15)
Let: the distance of the person from the streetlight the length of the shadow 15 ft 6 ft x
y
Using the above drawing, we are 4 ft/sec. looking for when Note that:
Comparing the small rectangle to the large triangle, from Geometry: 15
6
∙
So, 15
6
∙4
8
3
.
/
6 2
3
AnswerB 16) A particle moves along the x-axis so that its position at any time t (in seconds) is s (t )  2t 2  3t  5 . The
acceleration of the object at t = 2 seconds is:
(A) 19 units/s
2
(B) 11 units/s
2
(C) 16 units/s
Acceleration is the second derivative of the position curve (the first derivative is velocity). 2
3
4
5 3 (D) 4 units/s
2
AnswerB
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(E) 0 units/s
2
Note that the acceleration is the same for all values of . So, 4 2
/
APCalculus
APPracticeTest# Free Response: Part I: No Calculator
2 x  1, x  2
17) (1981 AB) Let f be defined by f ( x )  
1 2
 2 x  k , x  2.
(A) For what value(s) of k will f be continuous at x = 2? Justify with the definition.
(B) Find the average rate of change of the function on the interval [1, 4].
(C) Find the instantaneous rate of change of the function at x = 4.
(A) Continuity: A function, , is continuous at a.
is defined, b. lim →
exists, and c. lim →
iff:
. In the problem given, this boils down to: 2: When 2
1
2
becomes: 5
(B) The average rate of change is the slope of the line connecting the endpoints of the interval:
1: 1
2 1
4: 4
1
4
2
Slope
11 3
4 1
1
3 3
11 (C) The instantaneous rate of change at curve at 4. For 2:
1
2
4 is the slope of the 3 ,
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APCalculus
APPracticeTest# Part II: Calculators may be used.
18) Let f be the function defined by the equation f ( x )  x 3  2 x 2  6 x .
(a) Find the equations of the lines tangent and normal to the graph at the point 2, 2 .
(b) Find the equation(s) of the line(s) tangent to the graph of f and parallel to the line
(c) At what value of x, if any, is the tangent line horizontal?
7
(A) Find the equations of the lines tangent and normal to the graph at the point 2,
2
3
6 6 4
Tangent Equation is: Normal Equation is: slope) 2
2
3∙2
2∙2
4∙2
6∙2
6
3.
2 . (using point‐slope form) (normal slope is opposite reciprocal of tangent Page 9 of 11
APCalculus
APPracticeTest# (B) Find the equation(s) of the line(s) tangent to the graph of and parallel to the line We want the points where the slope of the curve is (i.e., the slope of 7
3) 3
4
6
7 3
4
13
0 Use the quadratic formula to determine that: .
,
.
.
so the points are: ,
.
.
and ,
.
Note: to get the ‐values of the points, substitute the ‐values into the original equation: 2
6 Tangent Equation 1 (point‐slope form): .
.
Tangent Equation 2 (point‐slope form): .
.
(C) At what value of x, if any, is the tangent line horizontal?
The tangent line is horizontal when the derivative is zero. 3
4
6
0 Using the quadratic formula, 4
4
√
4 3
2∙3
4
6
√88
6
.
,
.
Page 10 of 11
7
3. APCalculus
APPracticeTest# 19)
25
(a)
1
25
2
∙
3
(b)
3
2 25
3
4 So, the tangent line has slope and goes through the point Tangent Equation is: (using point‐slope form) (c) Continuity: A function, , is continuous at a.
is defined, exists, and b. lim →
c. lim →
3
lim
→
lim
→
3
7
iff: 4 (from above) 4 25
3, 4 . left and right limits
4 Since the limits from the left and right are the same at and are equal to the value of , which exists, we conclude that is continuous at . Page 11 of 11