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Question
Show that the exponential function f : R → R, given by f(x) = ex, is one-one but not onto. What happens if the co-domain is replaced by`R0^+` (set of all positive real numbers)?
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Solution
f : R → R, given by f(x) = ex
Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
f(x)=f(y)
⇒ `e^x = e^y`
⇒ x = y
So, f is one-one .
Surjectivity:
We know that range of ex is (0, ∞) = R+
Co-domain = R
Both are not same.
So, f is not onto.
If the co-domain is replaced by R+, then the co-domain and range become the same and in that case, f is onto and hence, it is a bijection.
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