2006 Form B Solutions

1a) x-int: x = -1.37312;   A =                                         

 

 

b) Y1 = R(x) = f(x) + 2                                    c) l: slope = f ’(0) = -1/2; point x=0, y = 3                            

   Y2 = r(x) = 0 + 2                                               eq y = -1/2x + 3; l & f intersect at x = 3.38987      

V =                             A =                                                 

 

2 a) f ‘ is decreasing for 1.7 < x < 1.9, so f “ is negative, so f is concave down

 

     f ‘                        

                                         0--------1.77--------2.5--------3

                                        f ‘’    +               -              +

 

b) f ‘(x) = 0 when x = 0, 1.77245, 2.50663; f increases, decreases, increases so check x = 1.77245 & x = 3 (why 3?)

f(3) = f(0) +  = 5.57893;                  f(1.77245) = f(0) + =5.68

 

absolute max at x = 1.77245

 

 

c) f ‘(2) = -.45902;  f(2) = f(0) +  = 5.62342;

     y – 5.62342 = -.45902(x – 2)

 

 

3. a) f(4) = 1 means a = 1/16; f ‘(4) = 1 means 2a(4) = 1 means a = 1/8; not possible.

 

b) g(4) = 64c – 16/16 = 1 means c = 1/32; If c = 1/32, g’(4) = (3/32)(16) – 2/16(4) = 1

    (note you can’t skip the part where you check that g’(4) = 1)

 

c) g’(x) = (3/32)x2 – (2/16)x;g’(.5) = -.03906, so g is decreasing at x = .5 (graph g’ & zoom in a bit to see that g’ is not always positive).

 

d)  so 4n = k; , so n = 4 & k = 256;

    h(x) = x4/256 so h(0) = 0;h’(x) = 4x3/256 so h’(0) = 0 & h’(x) is positive for 0<x<4.

 

 

 

 

4.a)  f ‘(22) = (15-3)/(20-24) = -3 calories/min/min                                    

     

 

  b)  on [0,4]

 

      f ‘(x) = -3/4x2 + 3x                 f ‘          

      f ’’(x) = -1.5x + 3                   0--------2---------4

              0 = -1.5x + 3                  f ‘’   +            -

               x = 2                                    

       so f ‘(x) has a max at x = 2 on [0,4] as f ‘’ goes from + to –.                               

     

      f ‘(2) = 3, f ‘(x) = 1.5 on 12<x<16, so f increasing at greatest rate at x = 2             

                                                                                   

c)  calories

 

d) original avg rate was 132 cal/12 min or 11 cal/min, so c = 4 gives a rate of 15 cal/min           

                                                                         

 

 

5. b) substitute y = c & y’=dy/dx=0 into the diff eq. to get, 0=(c-1)2cos(πx), so c = 1.

 

    c) y’(y-1)-2 = cos(πx)

          -(y-1)-1 = (1/π)sin(πx) + C

          sub x=1 & y=0 to get C = 1;

        -1/(y-1) = (1/π) sin(πx) + 1

          y = 1 - π/(sin(πx)+π)     

 

 

6. a) is the distance the car traveled in feet between time 30 sec and 60 sec.

        Trap. Estimate = =185 ft (note that the width of each

        interval is not the same).

         

    b) = change in velocity between time 0 & 30 = -14 - -20 = 6 ft/sec (if acc. Is

        the “rate” function then velocity is the “amount function” & )

           

   c) Yes by IVT as -5 is between v(35) & v(50)  

   

 

   d) Yes  (v(25) – v(0))/(25-0)=0, so MVT guarantees a c between 0 & 25 with a(c) = v’(c) = 0.