If f : [ 0 , ) [ 0 , ) and f ( x ) = x 1 + x , then f is [2003]

Question

If $f : [0, \infty) \to [0, \infty)$ and $f (x) = \frac{x}{1 + x},$ then $f$ is [2003]

Smart Achievers · Accepted Answer

(2)

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

$∴$ $f' (x) = \frac{(1 + x) - x}{{(1 + x)}^{2}} = \frac{1}{{(1 + x)}^{2}} > 0,$ $\forall x$

$∴$ $f is an increasing function \Rightarrow f is one-one.$

Now, $D_{f} = [0, \infty)$

For range, let $\frac{x}{1 + x} = y$ $\Rightarrow x = \frac{y}{1 - y}$

Now, $x \geq 0 \Rightarrow 0 \leq y < 1$

$∴$ $R_{f} = [0, 1) \neq Co-domain,$

$∴$ $f is not onto.$

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