CPhill

Note that     x^3 - x^2 -21x + 45 =  (x + 5) (x - 3) (x - 3)^2

Multiply through by this factorization and we  have

x^2 - 21x -3 = A(x - 3)^2  + B( x + 5) ( x-3) + C( x + 5)

x^2 -21x - 3 =  A( x^2 - 6x + 9) + B( x^2 + 2x -15) + C(x + 5)

Equate coefficients on each  side

1 =  A + B    →    B =  1 - A      (1)

-21 = -6A + 2B + C     (2)

-3 = 9A -15B + 5C     (3)

Sub (1) into (2) , (3)

-21 = -6A + 2 ( 1 - A) + C     →  -21 = - 6A + 2 - 2A + C →         -8A + C  =  -23      (4)

-3  = 9A - 15 ( 1 - A) + 5C  →  -3  =  9A - 15 + 15A  + 5C   →      24A + 5C =   12      (5)

Multiply  (4)  through by (-5)    add to  (5)

40A - 5C = 115

24A + 5C =  12

____________

64 A      =  127

A = 127/64

This simplifies to   ( x + 2x - 2) / ( x -1)  =  (3x -2) / ( x -1)

We have a horizontal asymptote  at  y= (3x/ x)  =  3

The range is  [(-inf , 3) U ( 3 , inf)

I like your thinking  !!!

Note that  we can    first  add  f(1) + f(2005)  and we  get

log (1) / ( log (1) + log (2005) )  +  log (2005) / ( log (2005) + log (1) )  =   

[ log (1) + log (2005) ]  / [ log(1) +Log (2005)] =   1

Next  we pair  f(2) + f(2004) and we  have

log (2) / ( log 2 + log 2004)  + log (2004) /( log (2004) + log (2) ) = (log 2 + log 2004) /(log 2 + log 2004) = 1

Continuing this pattern....We end up with   (2005 - 1 ) / 2 =  1002  pairs that each  evalute to 1

And we  have one evaluation  which is    n = (2005 + 1) / 2 =  1003

The evaluation of   this  is      log (1003) /( log 1003 + log (2006 -1003))  = 

log (1003) / ( log 1003  + log 1003)  =  1* log (1003) / 2 *log ( 1003 )  =  1/2 =  .5

So.....the sum is   1002 + 1/2  =   1002.5

\([f(x) = \frac{2-x}{\log (2 - x)}\)

Only the denominator matters

We can't take the  log of a  non-zero  number

Therefore  the domain is  ( -inf , 2)   

The roots  are  5 and  -9/2

Can you take it from  there ????

We can construct 12  congruent  isosceles triangles....with sides of  1 and  apex angles  of 30°

The area =   12* (1/2) ( side)^2 * sin (30°)  = 12* (1/2) (1^2) * sin (30°)  = 6 ( 1 / 2)  = 3   units^2

ab - 6a + 5b  =  37      subtract 30 from both sides   (the product of -6 * 5)        

ab - 6a + 5b - 30  =  37 - 30       factor the left side, simplify the  right

(a + 5) ( b - 6)  =  7

Let a = 2  and b = 7

abs ( a - b)  =  abs ( 2 - 7)  =  abs (-5)  =   5  is the  smallest  value

P.S.  .......check  to  see that  both of these  values work in  the  original  equation !!!!

Area = 36a^8b^4

Height = ab

Area =  (1/2) AC ( height)

2 *Area / (height ) =  AC

2 (36a^8b^4) / (ab) =  AC

72 a^7 b^3  =  AC

This is a  little  difficult to  see

Here's a graph :

Notice  that   0   is  not  in the range......

We just subtract  2  from both sides.....