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I actually needed an explanation.... sorry....

The answer is 55

You still have not said the sum

No prob

Sorry....I  made a slight mistake...it should be

[  14 + (1/3)(14-2)  ,  4 + (1/3)(4  - - 2) ]  =

[ 14 + 4,  4 + 2  ]  =

(18,  6)

Which agrees with your answer

THX for the correction   !!!!

well the domain range is 

This is a 2nd degree quadratic polynomial so its graph is a parabola.

Its general form is y=ax2+bx+c where in this case a = 1 indicating that the arms go up, b = 7, c = - 18 indicating the graph has y-intercept at - 18.

The domain is all possible x values that are allowed as inputs and so in this case is all real numbers R .

The range is all possible output y values that are allowed and so since the turning point occurs when the derivative equals zero,
⇒2x+7=0⇒x=−72
The corresponding y value is then g(−72)=−1214

Hence the range {y∈R∣y≥−1214}=[−1214;∞)

I have included the graph underneath for extra clarity.

graph{x^2+7x-18 [-65.77, 65.9, -32.85, 32.9]}

Thank you Cphill 

Mmmmm....let's see

2^1024  - 1  =

(2^512  + 1) ( 2^512  - 1)

And

(2^512 - 1)  =   ( 2^256 + 1) (2^256 - 1)

And

(2^256 - 1)  = (2^128 + 1)(2^128 - 1)

So we see that  the pattern of the smaller factors is

(2^64 - 1)

(2^32 - 1)

(2^16 - 1)

(2^8 - 1)

(2^4 - 1) =  (2^2  + 1)  ( 2^2 -  1)  =   (5) (3)

(2^2 - 1) =   (2 + 1) (2 - 1)  =  (3) (1)    1 is not  a prime

So....the  smallest  prime factors  =   3 , 5

Their product  =15

We  have  the points on the  original graph  (-1,1)   ( 0,2)   and  (2,4)

The inverse reverses the  coordinates  ( 1, -1)  ( 2,0)  and  ( 4,2)

So

f ^-1 (1)   +  f ^-1 (2)  + f ^-1(4)    =    -1  +  0   + 2    =     1

Thanks for trying but the answer is (18,6)

Implcitly taking the derivative, we get y*y' + 3x^2 = 2*xy*y'.  Plugging in x = -1, we get y' = -3.

For x = -1, y = 2 and y = -1, so the tangent line are y + 1 = -3(x + 1) and y - 2 = -3(x + 1).

If  the line of symmetry  goes through  x  = -2.....then  this is the  x coordinate of the  vertex

So we have  this

y =  a  ( x - h)  +  k        where  h  = -2

We  need to solve  these two equations  for a and  c

1  =  a ( 5 + 2)^2  +  k   →     1 = 49a  + k   →  49a + k   =    1       (1)

-3  =  a(4 + 2)^2  +  k   →    -3  = 36a  + k    → 36 a + k =    -3      (2)

Subtract   (2) from (1)   and we have that

13a  =  4

a = 4/13

And 

  49(4/13)  + k   =  1

k =  1  - 196/13

k = -183/13

So we have

y =  (4/13)  ( x + 2)^2   - 183/13

13 y  =  4 (x^2  + 4x + 4)    -183

13y = 4x^2  + 16x  + 16  - 183

13y =  4x^2  + 16x - 167

y =  (4/13)x^2  + (16/13)x - 167/13

So

a  +  b +  c  =    ( 4 + 16 - 167)  /13   =   -147/13

To get the point that divides AB so that AP:PB = r, you multiply the coordiantes of A by r and the coordinates of B by 1 - r, and add them.  For example, for r = 1/2, you just take the average of the coordinates of A and the coordinates of B.

So for point C, we take r = 1 + 1/3 = 4/3.  So

C = 4/3 (2,-2) + (-1/3) (14,4) = (-2,-4).

The answer to this problem is actally 18,6

Thanks for trying but it is incorrect

64 pi  = pi * r^2       divide out  pi

64  =r^2      take the positive root

8  = r

The intersection points will be  where  the line with the equation  ( x - 3)^2 + ( y + 8)^2 = 12^2   intersects  the line  y  = 2x +1

Sub  the equation of the line into the equation of the  circle  for  y  and we have

(x - 3)^2  +  [ (2x + 1) + 8 ]^2  =  12^2      simplify

x^2  - 6x  +  9  +   ( 2x + 9)^2   =144  

x^2  - 6x + 9  +  4x^2  + 36x  +  81   = 144

5x^2  + 30x  + 90  = 144        subtract 144 from both sides

5x^2  + 30x  - 54  =  0

x =   -30  +  sqrt  ( 900 + 20 *54)              -30 + sqrt (1980)            -30 + 6sqrt (55)

       ______________________  =       ________________  =  ________________  =

                2 * 5                                                   10                                 10

[-15 + 3sqrt ( 55)]  / 5

And the other  x value  is    [-15  -3sqrt (55) ] / 5

So  y =    (2/5) ( -15  + 3sqrt (55)) + 1  =   -6  + (6/5)sqrt (55)  +1  =   -5 + (6/5)sqrt (55)

And  the other y  =  -5 - (6/5)sqrt (55)

So  the intersection points are

(  [ -15 + 3sqrt (55)] / 5, -5 + (6/5)sqrt (55)  )    and  ( [ -15 -3sqrt (55) ] / 5  , -5  - (6/5)sqrt (55) ]

Point on  f =  (3.4)

In g(x)   the "3"   compresses  the graph by a factor of 3

So.....the new x coordinate  =  3/3    =  1

And in g(x), the "1"  shifts the y coordinate of  f  up by 1  =  4 + 1  = 5

So   ( 1, 5)   lies on g(x)

New point  C = 

[  14 + (1/3)(14-2)  ,  2 + (1/3)(4  - - 2) ]  =

[ 14 + (1/3) (12)  ,    2  +  (1/3)(6)  ]  =

[ 14  + 4 , 2 + 2  ]  =

( 18 , 4)

   sorry....that would make the mean EXACTLY 5 min 50 sec....you'll need to subtract ONE more second from the 2100 to get the answer...