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#1 2006-10-24 20:20:31

dericteong
Member
Registered: 2006-10-16
Posts: 13

Help me on functions pls..

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#2 2006-10-25 04:09:10

Anon
Guest

Re: Help me on functions pls..

Not entirely sure if my answer is correct

i. (f - g)(x)
   (5 + 2/x) - (x - 1)
   4 - 1/x
   Then 1 is substituted in as x, seeing as x=1
   4 - 1/1
   3

ii. Not sure what the little circle stands for, but this MAY be the answer:
   5 + 2/(x - 1)
   Then substitute 2 because x=2
   5 + 2/(2 - 1)
   5 + 2/1
   7

iii. y = 5 + 2/x
    2/x = y - 5
    x = 2(y - 5)
   f^-1 = 2y - 10

#3 2006-10-25 05:32:48

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: Help me on functions pls..

Anon wrote:

Not entirely sure if my answer is correct

i. (f - g)(x)
   (5 + 2/x) - (x - 1)
   4 - 1/x

That should be 6 + 2/x - x = (2 + 6x - x^2)/x


Anon wrote:

iii. y = 5 + 2/x
    2/x = y - 5
    x = 2(y - 5)
   f^-1 = 2y - 10

That should be x = 2/(y-5) , f^-1 = 2/(x-5)

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#4 2006-10-28 20:09:54

mahmoudaljamel
Member
Registered: 2006-09-03
Posts: 18

Re: Help me on functions pls..

1. ( f - g ) (x)
  Normal subtraction as the subtraction of numbers
    ( f - g ) (x) = (5 + 2/x) - (x-1)
                     = 5 + 2/x -x + 1
                     = 6 + ( 2-x²)/x
    ( f - g ) (1) = 7

2. ( f o g ) (x) = f ( g (x) )
                     = f ( x-1)
                     = 5 + 2 / ( x-1)
   
    ( f o g ) (2) = 7

3. f-¹ (x)  :

               y = 5 + 2/x
            y-5 =2/x
       x (y-5) = 2
               x = 2 / ( y-5 )


If you always do what you always did, you'll always get what you always got

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