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 \(B\)?
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
Very good! Amazing, CPhill!
It was nothing....really !!!
+4622 
