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.\(The/ function /f/ satisfies/ f(\sqrt{x + 1}) = \frac{1}{x} /for/ all/x \ge -1, x\neq 0. Find f(2).\)

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

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

 

4.

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

 Apr 12, 2018

Best Answer 

 #1
avatar+7612 
+3

1.   The function  \(f\)  satisfies   \(f(\sqrt{x+1})=\frac1x\)   for all   \(x\geq-1\,,\quad x\neq0\) .   Find  \(f(2)\) .

 

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

 

\(\sqrt{x+1}=2\\~\\ x+1=4\\~\\ x=3\)

 

So......

 

\(f(\sqrt{x+1})=\frac1x\\~\\ f(\sqrt{3+1})=\frac13\\~\\ f(\sqrt{4})=\frac13\\~\\ f(2)=\frac13\)

.
 Apr 12, 2018
edited by hectictar  Apr 12, 2018
 #1
avatar+7612 
+3
Best Answer

1.   The function  \(f\)  satisfies   \(f(\sqrt{x+1})=\frac1x\)   for all   \(x\geq-1\,,\quad x\neq0\) .   Find  \(f(2)\) .

 

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

 

\(\sqrt{x+1}=2\\~\\ x+1=4\\~\\ x=3\)

 

So......

 

\(f(\sqrt{x+1})=\frac1x\\~\\ f(\sqrt{3+1})=\frac13\\~\\ f(\sqrt{4})=\frac13\\~\\ f(2)=\frac13\)

hectictar Apr 12, 2018
edited by hectictar  Apr 12, 2018
 #2
avatar+7612 
+3

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+100571 
+1

Very nice, hectictar  !!!!

 

 

cool cool cool

CPhill  Apr 12, 2018
 #4
avatar+7612 
+2

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+100571 
+1

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

 

 

 

cool cool cool

CPhill  Apr 12, 2018
 #6
avatar+7612 
+2

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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