Show that the exponential function f : R → R, given by f(x) = e^x, is one-one but not onto. What happens if the co-domain is replaced by $R \overset{+}{0}$ (set of all positive real numbers)?

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Solution

f : R → R, given by f(x) = e^x

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
f(x)=f(y)
$\Rightarrow e^{x} = e^{y} \Rightarrow x = y$
So, f is one-one.

Surjectivity:
We know that range of e^x 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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