AP CALCULUS AB
PRACTICE EXAM
1) Multiply by clever form of 1
3 and 1/3
2) Factor First then use direct substitution
3) Sub in 2 set equal to each other:
(2)2 – 3(2) + 9 = 2k + 1
7 = 2k + 1
6 = 2k
3=k
4) Chain Rule: Peel the onion
5)
f ¢(x) = 6x 2 - 6x -12
= 6(x - 2)(x +1)
F’ sign chart look for where it changes from
– to +
6) Chain rule
Then sub in 1
7) Multiply and divide by 2 & 1/2 –
u-substitution
8)
LRAM = 2 (4 + k + 8) = 2 (12 + k)
RRAM = 2(12 + 8 + k) = 2(20 + k)
(24 + 2k + 40 + 2k)/2 = 52
64 + 4k = 104
Work backwards
9)
s(t) = ò (4 - 6t 2 )dt
s(t) = 4t - 2t 3 + C
7 = 4(1) - 2(1)3 + C
C=5
s(t) = 4t - 2t 3 + 5
s(2) = 4(2) - 2(2)3 + 5
= -3
10)
ax 2 +12
f (x) = 2
x +b
Horizontal asymptote y = 3 a = 3 since debrees are the same
Vertical asymptote x = 2 How will x2 + b give you an answer of 2
11)Quotient Rule
(x +1)(-1e- x ) - e- x
y¢ =
(x +1)2
Sub in x = 1 right away- don’t simplify first
12) Work backwards antiderive
• 13) Foil First- then anitderive
• 14) on 2012 exam
• Just use the formal definition of derivative
f (x + h) - f (x)
lim
h®0
h
15) Think about what graph the slope field
would make
• 16) f’(x) is defined for all real numbers so f(x)
must be continuous, thus eliminating d
• Also f’(x) is always going to positive so f must
be always increasing- only choice E
17)Set equations equal to each other
x – 2x2 = -5x
2x2 – 6x = 0
2x(x – 3)
x = 0, x = 3
ò (x - 2x ) - (-5x)dx
3
2
0
3
Top curve- bottom curve
2
(-2x
+ 6x)dx
ò
0
2
- x 3 + 3x 2 from3to0
3
• 18) f’(1) = 2(1) + 1 = 3
• y – 4 = 3(x – 1)
• Y = 3x +1
• Y= 3(1.2) + 1
• Y = 3.6 + 1 = 4.6
• 19) f’(x) = 3x2-12
F’’(x) = 6x -12
6x – 12 = 0
X=2
Set up chart
20)g’(x) = 2x – 3
3
ò (2x - 3)dx
1
• 21) x – 2 = 20
•
x = 22
• 22) set up chart
SECTION 1
PART B
76)A particle moves along the x-axis so that any
time t > 0 its velocity is given by
v(t) = t2ln(t + 2). What is the acceleration of
the particle at time t = 6?
Could do by hand or use calculator:
A(t) = v’(t)
Calculator:
By Hand:
Enter v(t) on calculator
Use dy/dx key and evaluate at x = 6
A(t) = 2t ln (t + 2) + t2 ( 1 )
t+2
Evaluate a(6)
77) If
3
5
0
3
ò f (x)dx = 6 and ò f (x)dx = 4 then ò (3 + 2 f (x))dx =
5
5
ò (3 + 2 f (x))dx
0
5
5
0
0
= ò 3 + 2 ò f (x)dx
= 15 + 2(10)
= 15 + 20
= 35
0
• 78) Derivative is the RATE of change in the
temperature
• 79) The rate of change in the water is given by
the equation
• Y = 9sin(√(x + 1)
• Enter into calculator and evaluate the integral from 0
min to 6 min.
• Get 45.031019
Subtract from 81.637
80)Since graph is increasing from left to right,
Left Sum is an underestimates.
Trapezoidal underestimates when graph is
concave down. (overestimate when graph is
concave up)
Show graph
• 81) The first derivative of the function f is
given by
f ¢(x) = x - 4e
- sin(2x)
• How many points of inflection does the graph
of f have on the interval 0 < x < 2pi ?
f ¢¢(x) = 1 - 4e- sin(2 x ) (- cos(2x))2
= 1 + 8 cos(2x)e- sin(2 x )
Graph on calculator setting window from 0 to 2pi and find how many times it crosses the
x-axis (Where it is equal to 0)
Note: Could also look at the graph of the first derivative and see how many mins and maxs
there are
• 82) Think of a horizontal line or a cubic curve
• Only c is true has greater than or equal to
• D) is what Theorem?
• E) is what theorem
83)
x
f(x)
2.5
31.25
2.8
39.20
3.0
45
3.1
48.05
Find the rate of change between 2.8 and 3
(45 – 39.2)/(3 – 2.8) = 29
Find the rate of change between 3 and 3.1
(48.05-45)/(3.1-3.0) = 30.5
The only value it could be is 30 between those two.
• 84) Where the first derivative is negative
(below the x-axis) is where the graph can be
decreasing Answer E
• Where is the graph increasing?
• 85) Skip Didn’t do cross perpendicular
sections
• 86 & 87 Use table
88)The rate at which water is sprayed on a field
of vegetables is given by R(t) = 2 1 + 5t 3
where t is in minutes and R(t) is in gallons per
minute. During the time interval 0 < t < 4,
what is the average rate of water flow, in
gallons per minute?
b
Average Value:
1
f (x)dx
ò
b-a a
Set up and solve using your calculator
• 89)
• h(x) = (2f(x) + 3)(1 + g(x))
• Hint: FOIL, Derive, and then substitute
• h(x) =2f(x) + 2f(x)g(x) + 3 + 3g(x)
H’(x) = 2f’(x) + 2f’(x)g(x) +2f(x)g’(x) + 0 + 3g’(x)
H’(1) = 2(-2) + 2(-2)(-3) + 2(3)(4) + 3(4)
90)
f & g are ____Inverse_______ functions
Set up a table- slopes are reciprocals (not
opposite reciprocals)
• 91) A particle moves along the x-axis so that its
velocity at any time t > 0 is given by
-t
• v(t) = 5te -1 . At t = 0, the particle is at
position x = 1. What is the total distance traveled
by the particle from t = 0 to t = 4?
Take the absolute value and integrate from 0 to 4
4
ò 5te
0
-t
-1
92)
Calculator method
Enter y = sin(x3) in y = screen
and use zero key
Make sure do left and right bound between
0 and 2
OR Use solver key