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.The/function/f/satisfies/f(x+1)=1x/for/all/x1,x0.Findf(2).

2.Let/f(x)=3x24x.Find/the/constant/k/such/that/f(x)=f(kx)for/all/real/numbers/x.

3.Find/all/complex/numbers/z/such/that/z2=2i.Write/your/solutions/in/a+bi/form/,separated/by/commas./So,/"1+2i,3i"/is/an/acceptable/answer/format,/but/"2i+1;i+3"/is/not./(Dont/include/quotes/in/your/answer.)/Note:/This/problem/is/not/about/functions.

 

4.

Let/f/be/a/function/such/that/f(x+y)=x+f(y)/for/any/two/real/numbers/x/and/y/.Iff(0)=2,then/what/is/f(2012)?

 Apr 12, 2018

Best Answer 

 #1
avatar+9482 
+4

1.   The function  f  satisfies   f(x+1)=1x   for all   x1,x0 .   Find  f(2) .

 

We want to find an  x  value such that...

 

x+1=2 x+1=4 x=3

 

So......

 

f(x+1)=1x f(3+1)=13 f(4)=13 f(2)=13

 Apr 12, 2018
edited by hectictar  Apr 12, 2018
 #1
avatar+9482 
+4
Best Answer

1.   The function  f  satisfies   f(x+1)=1x   for all   x1,x0 .   Find  f(2) .

 

We want to find an  x  value such that...

 

x+1=2 x+1=4 x=3

 

So......

 

f(x+1)=1x f(3+1)=13 f(4)=13 f(2)=13

hectictar Apr 12, 2018
edited by hectictar  Apr 12, 2018
 #2
avatar+9482 
+4

2.   Let   f(x)  =  3x2 - 4x  .  Find the constant  k  such that   f(x)  =  f(k - x)   for all real numbers  x .

 

f(x)   =   f(k - x)

                                                 And   f(x)  =  3x2 - 4x

3x2 - 4x   =   f(k - x)

                                                 And   f(k - x)  =  3(k - x)2 - 4(k - x)

3x2 - 4x   =   3(k - x)2 - 4(k - x)

 

3x2 - 3(k - x)2 - 4x + 4(k - x)   =   0

 

3[x2 - (k - x)2] - 4[x - (k - x)]   =   0

 

3[x + (k - x)][x - (k - x)] - 4[x - (k - x)]   =   0

 

3[x + k - x][x - k + x] - 4[x - k + x]   =   0

 

3[ k ][ 2x - k ] - 4[ 2x - k ]   =   0

 

( 2x - k )( 3k - 4 )   =   0

 

2x - k  =  0      or      3k - 4  =  0

 

k  =  2x                      k  =  4/3

 

The constant value that works is   k  =  4/3

 Apr 12, 2018
edited by hectictar  Apr 12, 2018
edited by hectictar  Apr 12, 2018
edited by hectictar  Apr 12, 2018
 #3
avatar+130327 
+2

Very nice, hectictar  !!!!

 

 

cool cool cool

CPhill  Apr 12, 2018
 #4
avatar+9482 
+3

Ah...thanks..but....my final answer happened to be right but my working out was all messed up!!  blush I had to fix it!!

hectictar  Apr 12, 2018
 #5
avatar+130327 
+2

I don't know....you're always pretty good with these function problems.....!!!

 

 

 

cool cool cool

CPhill  Apr 12, 2018
 #6
avatar+9482 
+3

4.   Let  f  be a function such that   f(x + y)  =  x + f(y)   for any two real numbers  x  and  y .

      If  f(0)  =  2  ,  then what is  f(2012)  ?

 

f(x + y)  =  x + f(y)

 

f( -2012 + 2012 )   =   -2012 + f(2012)

 

f(0)   =   -2012 + f(2012)

 

2   =   -2012 + f(2012)

 

2 + 2012   =   f(2012)

 

2014   =   f(2012)

 Apr 12, 2018

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