If f:[0, infty) to [0, infty) and f(x) = frac | vedclass

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(B) Given $f(x) = \frac{x}{1+x}$ for $x \in [0, \infty)$.To check for one-one: Let $f(x_1) = f(x_2)$,then $\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$.$x_1

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One-one but not onto

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(B) Given $f(x) = \frac{x}{1+x}$ for $x \in [0, \infty)$.To check for one-one: Let $f(x_1) = f(x_2)$,then $\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}$.$x_1(1+x_2) = x_2(1+x_1) \implies x_1 + x_1x_2 = x_2 + x_1x_2 \implies x_1 = x_2$.Thus,$f$ is one-one.To check for onto: The range of $f(x) = \frac{x}{1+x} = \frac{1+x-1}{1+x} = 1 - \frac{1}{1+x}$.As $x$ ranges from $0$ to $\infty$,$1+x$ ranges from $1$ to $\infty$,so $\frac{1}{1+x}$ ranges from $1$ to $0$.Therefore,$f(x)$ ranges from $1-1=0$ to $1-0=1$,i.e.,$[0, 1)$.Since the codomain is $[0, \infty)$ and the range is $[0, 1)$,the range $
eq$ codomain.Thus,$f$ is not onto.Therefore,$f$ is one-one but not onto.

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

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