The function f: R rightarrow R defined by f(x | vedclass

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(C) To check for injectivity, let $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$.$\frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}}$Squaring both sides:

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injective but not surjective

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(C) To check for injectivity, let $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$.$\frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}}$Squaring both sides:$\frac{x_1^2}{1+x_1^2} = \frac{x_2^2}{1+x_2^2}$$x_1^2(1+x_2^2) = x_2^2(1+x_1^2)$$x_1^2 + x_1^2x_2^2 = x_2^2 + x_1^2x_2^2$$x_1^2 = x_2^2$Since $f'(x) = \frac{1}{(1+x^2)^{3/2}} > 0$ for all $x \in R$, the function is strictly increasing.Therefore, $f(x_1) = f(x_2) \implies x_1 = x_2$. Thus, $f$ is injective.To check for surjectivity, let $y = \frac{x}{\sqrt{1+x^2}}$.Since $x^2 Thus, the range of $f$ is $(-1, 1)$, which is not equal to the codomain $R$.Therefore, $f$ is not surjective.Hence, the function is injective but not surjective.

The function $f: R \rightarrow R$ defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$ is

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