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avatar+3821 

Let  \(f(x) = Ax - 2B^2\)and \(g(x) = Bx\), where \(B \neq 0\) . If \(f(g(1)) = 0\) , what is \(A\)  in terms of \(B\)?

 Feb 5, 2018
 #1
avatar+95829 
+2

f(x)  =   Ax - 2B^2      g(x)   =  Bx      

 

g(1)  =  B(1)  = B

 

So

 

f( g(1) )  =  f(B)   =  0

 

A(B) -  2B^2  =  0

 

A (B)  =  2B^2       

 

A  =    2B

 

 

cool cool cool

 Feb 5, 2018
 #2
avatar+3821 
+1

Very good! Amazing, CPhill!

 Feb 5, 2018
 #3
avatar+95829 
0

It was nothing....really    !!!

 

 

cool cool cool

 Feb 5, 2018

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