Solutions Solutions Solutions Solutions Solutions Solutions Solutions

Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Worksheet # 104 Mixed Review 1

“Graph”of f Terrace pt Local Min

Some “y” value Local Min

2 4 5 Since f’(x) > 0 until 2 and then 4; the graph of f must be increasing until 4, and decreasing thereafter Therefore the answer is D 0 and 5 2

f’’<0

3

4

f’’undefined f’’>0 f’’<0 f’’=0

Since f’’(x) changes from positive to negative or from negative from positive there are inflection points at D) 2 and 3 f (b) − f (a) f '( c ) = b−a (e − 1) − (1 − 0) e c − 2c = = e − 2 ⇒⇒⇒ ec − 2c − e + 2 = 0 ⇒ Calculate the zero ⇒ c=.351 (1 − 0)

y = e x ⇒⇒ x = ln ( y ) 2

π ∫ ( ln( y ) ) dy ≈ 0.592 2

1

5

xy 2 − 2 y + 4 y 3 = 6 ⇒ y = 1 → x = 4 y2 (

dx dy dy dy dy = 0 ⇒⇒ (2 xy − 2 + 12 y 2 ) = − y 2 ) + 2 xy − 2 + 12 y 2 dx dx dx dx dx

− y2 −(1) 2 −1 dy = = @(4,1) = 2 dx (2 xy − 2 + 12 y ) 8 − 2 + 12 18

6

v(t ) = 3t 2 + c ⇒ v(0) = 1∴1 = 0 + c ⇒⇒ v(t ) = 3t 2 + 1 s (t ) = t 3 + t + c ⇒ s (0) = 3 ∴ 3 = 0 + 0 + c ⇒⇒ s (t ) = t 3 + t + 3

7

f ( x 2 )′ = f ′( x 2 ) • 2 x since f ′( ) = 5

− 1 ⇒ f ′( x 2 ) = 5 x 2 − 1 ⇒⇒ 2 x 5 x 2 − 1

Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Worksheet # 104 Mixed Review

8

k

k

k

x3 k3 sin cos cos ( 1) xdx = x ⇒⇒ − x = ⇒⇒ − k − − = ∫0 ∫0 0 3 0 3 k

2

k3 + 1 = 0 CALCULATE ZERO ⇒≈ 1.300 3 Try each choice: A)2 x sin x 2 B) − 2 x sin x 2 − cos k −

9

2 x sin x 2 • 2 x − (− cos x 2 )(2) C) 4 x2 10

D)2cos x 2 + 2 x(−2 x sin x 2 )

ln(.9) 9 ln(.9) (t ) ln(.4) ⇒ ln(.4) = 9 ⇒ t = ln(.9) ≈ 78

y = ne kt ⇒⇒ 45 = 50e9 k ⇒ .9 = e9 k ⇒ ln(.9) = 9k ⇒ k = 20 = 50e

ln(.9) (t ) 9

⇒ .4 = e

ln(.9) (t ) 9

9

11

∴E

X=”10” Y=”24” Dx/dt=3 Dy/dt= ? ft/sec x2 + y 2 = z 2

(t )

Z=16 Dz/dt=0

dx dy dz dy dy −5 ft + 2y = 2 z ⇒⇒ 10(3) + 24 = 0 ⇒ = dt dt dt dt dt 4 sec 2nd Fundamental Theorem of Calculus:∴

2x

12

1 2 •2= 3 (1-(2x) ) (1-8x 3 ) 13

Given: v(5) = 0 5

∫ a(t)dt = v(5) - v(0) = -π +3 (Area under curve- 1/4 circle w/radius 2 & triangle) 0

0 − v(0) = -π +3 ⇒⇒ v(0) = π − 3

14

15

⎛t⎞ 5 ∫ tan −1 ⎜ ⎟ dt ≈ 124 ⎝5⎠ 0 Use Calculator to find y’(2)

ft sec

24

y′(2) ≈ −2.0806; y(2) = 0 ∴ y-0=-2.0806(x-2) y-intercept: let x= 0 ⇒⇒ y ≈ 4.161

Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Worksheet # 104 Mixed Review