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3)     a) f '(x) = 6x - 5,  f '(-1) = 6(-1) - 5 = -11

b)  f '(x) = 2x - 1/x² ,     f '(-1) = 2(-1) - 1/(-1)² = -3

c)  f(x) = x^1/5 ,   f '(x) = x^-4/5/5 = 1/5x^4/5 ,     f '(-1) = 1/(-1)^4/5 = 1/5

d)  f '(x) = 2x² - 3/x² + 10/x³ ,     f '(-1) = 2 - 3 - 10 = -11

e)  set f '(x) = 0,     6x - 5 = 0

6x = 5

x = 5/6

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4a)  Zeros of f(x) = x(x² - 3) = 0 ,     zeros at 

     f '(x) = 3x² - 3 = 3(x² - 1) = 3(x+1)(x-1) = 0,  critical points at (1,-2), (-1,2)

Min     Max
 

x	f '(x)
x < -1	+
x = -1	0
-1	-
x = 1	0
x > 1	+

 

4b)  zeros of f(x) = x³ - 1 = 0
x³ = 1

  x = 1

    zeros of f '(x) = 3x² = 0,  x = 0 ,  critical point at (0, -1)   inflection point

 

x	f '(x)
x<0	+
x=0	0
x>0	+

 

5)  A(x) = x(24 - 2x) = 24x - 2x²
A'(x) = 24 - 4x = 0
 -4x = -24

    x = 6

Therefore, the amount to be turned up is 6 meters.

6) 
A(x) = x(900 - 3x)/2 = 450x - 3x²/2
A'(x) = 450 - 3x = 0

                  450 = 3x

                     x = 150

Therefore, the maximum area is (150)(225) = 33,750

7)  Let x = the number of nickels to raise the rates

# of passengers	rate per day in cents
2000 - 200x	20 + 5x

I(x) = (2000 - 200x)(20 + 5x) = 40000 +10000x - 4000x - 1000x²                               = 40000 + 6000x - 1000x² I'(x) = 6000 - 2000x     0 = 6000 - 2000x 2000x = 6000     x = 3

Therefore, they need to raise the rate by 3 nickels to 35 cents

        8)     a)  v(t) = 32t - 128

b)  v(3) = -32, moving backward,     v(4) = 0, not moving,     v(5) = 32 moving forward

c)  a(t) = 32

d)  All times yield a constant acceleration of 32.

9)     a)  v(t) = 256 - 32t
b)  v(6) = 64,  v(8) = 0,  v(10) = -64

c)  a(t) = -32

d)  0 = 256t - 16t²

     0 = 16t(16 - t)  therefore, t = 0 or t = 16.  t = 0 is before you hit it.  therefore, it will hit the ground at t = 16.

e)  v(16) = 256 - 32(16) = -256

f)  0 = 256 - 32t

    32t = 256

    t = 8  Therefore, it will take a time of t = 8

g)  h(8) = 256(8) - 16(8)² = 1024

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