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

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

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Injective but not surjective

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(C) To check for injectivity,let $f(x_1) = f(x_2)$.$\frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}}$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 \Rightarrow x_1 = x_2$ (since the function is strictly increasing,$x_1 = x_2$).Thus,$f$ is injective.To check for surjectivity,let $y = \frac{x}{\sqrt{1+x^2}}$.$y^2 = \frac{x^2}{1+x^2} \Rightarrow y^2(1+x^2) = x^2 \Rightarrow y^2 = x^2(1-y^2) \Rightarrow x^2 = \frac{y^2}{1-y^2}$.For $x$ to be a real number,$1-y^2 > 0$,which means $y^2 The range of $f$ is $(-1, 1)$,which is not equal to the codomain $R$.Therefore,$f$ is not surjective.

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

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