Functional fun
The domain of a function is all the real numbers. It is also known that for all real values of , and .
Is it necessarily true that ?
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1 solution
The given conditions may be written as f ( y 2 + 5 ) = ( y 2 + 5 ) 2 − 2 5 ; f ( y 2 + 3 ) = ( y 2 + 3 ) 2 − 2 5 . This seems to suggest that in general, f ( x ) = x 2 − 2 5 , and in therefore f ( 1 ) = 1 2 − 2 5 = − 2 4 . It is therefore quite possible that f ( 1 ) = − 2 4 .
In fact, for every x ≥ 3 , it must be true that f ( x ) = x 2 − 2 5 . Proof: Let x ≥ 3 and define y = x − 3 . Then f ( x ) = f ( y 2 + 3 ) = y 4 + 6 y 2 − 1 6 = ( x − 3 ) 2 + 6 ( x − 3 ) − 1 6 = ( x 2 − 6 x + 9 ) + ( 6 x − 1 8 ) − 1 6 = x 2 − 2 5 . However, no such reasoning applies for x < 3 . We know absolutely nothing about the behavior of the function on that interval; it need not even be continuous or expressible by an equation! Therefore it is quite possible that f ( 1 ) = − 2 4 ; its value may be any real number.