The function f:R to left[ { - frac{1}{2},frac | vedclass

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(D) Given $f:R 	o \left[ { - \frac{1}{2},\frac{1}{2}} ight]$ defined by $f(x) = \frac{x}{1 + x^2}$.First,check for injectivity:$f'(x) = \frac{(1 +

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surjective but not injective

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(D) Given $f:R 	o \left[ { - \frac{1}{2},\frac{1}{2}} ight]$ defined by $f(x) = \frac{x}{1 + x^2}$.First,check for injectivity:$f'(x) = \frac{(1 + x^2)(1) - x(2x)}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2} = \frac{(1 - x)(1 + x)}{(1 + x^2)^2}$.Since $f'(x)$ changes sign at $x = 1$ and $x = -1$,the function is not monotonic,hence it is not injective.Next,check for surjectivity:Let $y = \frac{x}{1 + x^2}$. Then $yx^2 - x + y = 0$.For $x$ to be a real number,the discriminant $D \ge 0$.$D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \ge 0$.$4y^2 \le 1 \Rightarrow y^2 \le \frac{1}{4} \Rightarrow y \in \left[ -\frac{1}{2}, \frac{1}{2} ight]$.Since the range is equal to the codomain,the function is surjective.Therefore,the function is surjective but not injective.

The function $f:R \to \left[ { - \frac{1}{2},\frac{1}{2}} \right],$ defined as $f(x) = \frac{x}{1 + x^2}$ is

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