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Question Number 119815 by bemath last updated on 27/Oct/20
Givenf(x+y)=4f(x).f(y)forallrealnumbersxandy.Iff(3)=32thenf(1)=_
Answered by Olaf last updated on 27/Oct/20
f(1+1)=4f(1)f(1)f(2)=4f(1)2f(2+1)=4f(2)f(1)f(3)=16f(1)3=32f(1)=23
Commented by bemath last updated on 27/Oct/20
yes....
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