First Principles AI

Solution

Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be defined by $f(x)=e^x$ .

Assume $f(x_1)=f(x_2)$ :

e^{x_1}=e^{x_2}

Taking $\ln$ on both sides:

\ln(e^{x_1})=\ln(e^{x_2})\Rightarrow x_1=x_2

So $f$ is one-to-one.

For onto, given $y\in\mathbb{R}^+$ , we need an $x\in\mathbb{R}^+$ such that $e^x=y$ .

Solving gives $x=\ln y$ .

But if $0\lt y\lt 1$ , then $\ln y\lt 0$ , which is not in $\mathbb{R}^+$ .

So with co-domain $\mathbb{R}^+=(0,\infty)$ , the function is not onto. Its range for $x>0$ is $(1,\infty)$ .

\boxed{f \text{ is one-to-one but not onto }\mathbb{R}^+\ (\text{so not bijective})}

If the intended co-domain is $(1,\infty)$ , then $f$ becomes bijective.