Summer Review - AP Calculus BC

I. Polynomials 1) a)

( ) b) x2 + 2x + 4 ; x ̸= 2 c)

1 ; x ̸= 4 x+1 5

−1 ; x ̸= 5 x+5

d)

x−8 ; x ̸= 4 x−4

7

2) x 2 + x4 − x 2 3)

1 31 (−1)2 02 12 22 32 42 + + + + + = (1 + 0 + 1 + 4 + 9 + 16) = 2 2 2 2 2 2 2 2

4)

( ) (a) x−5 7 + 4x2 + 2x8 (b) 2x− 4 (x + 1) (−3x + 4) 7

II.

Functions

1) Yes on the domain [−2, 2]. 1 √ 2) √ x+9+h+ x+9

3)

D: (−∞, −1) ∪ (−1, 4) ∪ (4, ∞)

D: (−∞, −5) ∪ (−5, ∞)

1

D: (−∞, −2) ∪ (−2, 3) ∪ (3, ∞)

D: R, R: (−∞, 9] ∪ 10

D: R, R: (−∞, −2) ∪ [0, 5]

4)

D: R (reals), R: Z (integers) D: R, R: integer multiples of 3

D: R, R: Z

5)

6)

7)  2  x − 1, x ≤ −1 f (x) = 1 − x2 , −1 < x < 1   2 x − 1, x ≥ 1

III. More with Functions 1) x+1 , x ̸= 0, 1 x(x − 1) 1+x f ◦g = , x ̸= 0, 1 Make sure you understand the 0. 1−x x+1+x−1 = x, x ̸= 1 f ◦f = x + 1 − (x − 1) f ·g =

2

(1) (2) (3)

2) √ f ◦ g = x2 − 4, D : (−∞, 2] ∪ [2, ∞) g ◦ f = x − 4, D : [2, ∞) √(√ ) f ◦f = x − 2 − 2, D : [6, ∞)

(4) (5) (6)

3) √ x; g(x) = x2 − 4 √ g(x) = x2 f (x) = x − 4; √ f (x) = x2 − 4; g(x) = x f (x) =

(7) (8) (9)

4) odd 5) odd 6)   x < −2 4, f (x) = −2x, −2 ≤ x ≤ 2   −4, x>2 odd

7) Proof. If you have questions, ask Dr. Horner. 8) g(x) = 2x − 3, −2x + 3 9) a) (f + h)(1) = 13 b) (k − g)(5) = 30 −

√

e) f −1 (x) = {(5, 3), (4, 2), (7, 1)} 2

f) g −1 (x) = x2 + 3, x ≥ 0

c) (f ◦ h)(3) = 4 √ d) (g ◦ k)(7) = 51

g)

10) −B < A < B 11) A < −B OR A > B 3

1 f (x)

=

{(

) ( ) ( )} 3, 51 , 2, 14 , 1, 71

(10)

12)

( (−∞, −1) ∪

) 7 ,∞ 3

13) −1 < x < 4

IV.

Word Problems and Proofs

1 s √ 1) T = √ √ L; 1 2 ft 2) cm2 2 (a) 36 2 πt s (b) 576 cm2 π 3) 120 m × 60 m 4) 2.21 cm × 2.21 cm 5) Proof sketch: Show statement is true for n = 1. Show that if statement is true for n, then it is true for n + 1. Therefore we can step-by-step prove for any n that the statement is true.

V.

Logs and Exponents

1) a)

√

x x

=

√1 x

f) log3

(1) 3

= −1

c) e1+ln(x) = ex

g) log1/2 (8) = −3 ( ) h) ln 12 = − ln(2)

d) ln(1) = 0

i) e3 ln(x) = x3

e) ln (e7 ) = 7

j)

b) eln(3) = 3

VI.

4xy −2 12x

−1 3

y −5

=

x4/3 y 3 3

Point-slope form

17 3 1) y = − x + 2 2 √ 2) y = x + 2 2

4

k) 272/3 = 9 ( )( ) 13 l) 5a2/3 4a3/2 = 20a 6 ( )3/2 m) 4a5/3 = 8a5/2 n)

3(n+2)! 5n!

=

3(n+2)(n+1) 5

VII.

MEMORIZED unit circle values

1) a) sin(0) = 0 √ ( ) 3 b) cos 7π = − 6 2 ( ) c) sec 2π = −2 3 ( ) d) sin π2 = 1

e) cos

(π) 3

=

( ( )) i) cos sin−1 21 =

1 2

√

3 2

j) cos(π) = −1 ( ) k) tan π6 = √13 ( ( )) l) sin−1 sin 7π = − π6 6

f) cos(0) = 1 ( ) √2 g) sin 3π = 2 4 ( ) h) tan 7π = −1 4

2) a) sin

( 17π ) 3

b) tan

( 43π )

c) cos

( 97π )

d) sin

=−

4

(

− 55π 3

3 2

( ) f) cot − 213π = −1 4

√

( ) =2 g) sec − 137π 3

3 3

= )

√

√

=

6

( ) e) cos − 71π = 6

√ 3 2

2 2

=−

√

3 2

h) csc

( 1000π ) 6

=

√2 3

)) ( ( i) tan−1 cot − 13π = − π4 4 ( ( √ )) j) sin cot−1 − 3 = 12 ( ( √ )) k) csc cos−1 − 23 =2 l) sin−1

(√ ) 3

VIII. Algebra 1) d) x = a) x = −6, 3 b) x = −1, 1 c) x = 2, 8

IX.

√ −5± 89 4

i) x = − 32

e) (−∞, −3) ∪ (3, ∞)

j) 5 Only!

f) [−3, 5] k)

g) x = −1, 2, 12 h) (−11, −3) ∪ (2, 14) ∪ (14, ∞)

Trig

1) (a) sin(A + B) = sin(A) cos(B) + cos(A) sin(B) (b) cos(A + B) = cos(A) cos(B) − sin(A) sin(B) (c) sin(2A) = 2 sin(A) cos(A) (d) cos(2A) = cos2 (A) − sin2 (A) cos(2A) = 2 cos2 (A) − 1

5

l)

ln(5) 3 ln(y)+3 2

64 4

= undefined

cos(2A) = 1 − 2 sin2 (A) ( ) √ 1 1 − cos(A) (e) sin A = 2 2 ( ) √ 1 1 + cos(A) A = (f) cos 2 2 2 2 (g) sec (A) = 1 + tan (A) (h) csc2 (A) = 1 + cot2 (A) 2) 4th 3) tan2 (θ) + 1 = sec2 (θ) and 1 + cot2 (θ) = csc2 (θ)

X.

Binomial Theorem

If the binomial theorem needs refreshing, please refer to the file binomial.pdf. 1) a8 + 8a7 b + 28a6 b2 + 56a5 b3 + 70a4 b4 + · · · 2) 4096x24 + 24576x22 y 4 + 67584x20 y 8 + 112640x18 y 1 2 3) −489888x4 4) 924 5) −8064

XI.

Geometric Series

1) −81, −27, −9, . . . 2) −3 3)

600 11

√ 9+3 3 4) 2 5)

137 111

6

XII.

Polar Coordinates

1) ( ( ) ( ) 7π 3π π) , 2, 2) −2, , 2, − 4 4 4 3)

(a)

(b)

(c)

7

XIII. Parametric 1)

(a)

(b) (c)

XIV.

Approximation

1) 0.02 2) 0.0248 NOT 0.02516 Make sure you know why.

XV.

Redefining Trig Functions

1) a) sin

(π)

b) tan

4

=

(π) 4

1 2

g) tan (−270°) = undef ined h) csc (1260°) = csc (180°) = undef ined

=1

( 3π )

= − 21 ) ( 3π ) ( d) csc − 19π = csc − 4 = −2 4 c) cos

i) cos (43.721 98°) =

4

e) sin (30°) =

j) sin (−948.6671°) =

1√ 1+ 3

0.956 1+0.956

k) sin2 (315°) + cos2 (315°) =

f) cos (120°) = − cos (60°) = − 1+1√3

XVI.

1 1+0.956

1 2

l) at n π2 , n integer

Unit Circle Graph

1) partial solution at http://commons.wikimedia.org/wiki/File:Unit_circle_angles.svg

8