If f:[0, infty) rightarrow[0, infty) is defin | vedclass

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(B) We have $f(x) = \frac{x}{1+x}$ where $f: [0, \infty) \rightarrow [0, \infty)$. For one-one: Let $f(x_1) = f(x_2)$. $\frac{x_1}{1+x_1} = \frac{x_2}

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one-one but not onto

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(B) We have $f(x) = \frac{x}{1+x}$ where $f: [0, \infty) ightarrow [0, \infty)$. For one-one: Let $f(x_1) = f(x_2)$. $\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$ $x_1(1+x_2) = x_2(1+x_1)$ $x_1 + x_1x_2 = x_2 + x_1x_2$ $x_1 = x_2$. Since $f(x_1) = f(x_2) \implies x_1 = x_2$, the function is one-one. For onto: Let $y = f(x) = \frac{x}{1+x}$. $y(1+x) = x \implies y + xy = x \implies y = x(1-y) \implies x = \frac{y}{1-y}$. Since $x \in [0, \infty)$, we must have $\frac{y}{1-y} \geq 0$. This implies $y \in [0, 1)$. The codomain is $[0, \infty)$, but the range is $[0, 1)$. Since Range $
eq$ Codomain, the function is not onto. Thus, $f$ is one-one but not onto.

If $f:[0, \infty) \rightarrow[0, \infty)$ is defined by $f(x)=\frac{x}{1+x}$, then $f$ is

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