1.

Let `f: RvecR`be such that `f(x)=2^x`. Determine:Range of `f`(ii) `{x :f(x)=1}`(iii) Whether `f(x+y)=f(x)f(y)`holds.

Answer» (i) `f(x)=2^(x)gt0` for every `x""inR`.
If `x""inR^(+)`, there exists `log_(2)x` such that `f(log_(2)x)=2^(log_(2)x)=x`.
`:."range "(f)=R^(+)`.
(ii) `f(x)=1implies2^(x)=1=2^(0)impliesx=0."So,"{x:f(x)=1}={0}`.
(iii) `f(x+y)=2^(x+y)=2^(x)xx2^(y)=f(x).f(y)`.


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