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

We are given the real-valued function

$$f:\; \mathbb R \;\rightarrow\; \left(-\dfrac12,\dfrac12\right), \qquad f(x)=\dfrac{x}{1+x^{2}}.$$

First we analyse the possible output values (the range) of the function, because this will tell us whether every element of the stated codomain $$\left(-\dfrac12,\dfrac12\right)$$ is actually attained.

To locate maximum and minimum values we differentiate. For a quotient $$\dfrac{u(x)}{v(x)}$$ the quotient rule says

$$\left(\dfrac{u}{v}\right)'=\dfrac{u'v-u\,v'}{v^{2}}.$$

Here $$u(x)=x$$ and $$v(x)=1+x^{2}$$, so $$u'(x)=1$$ and $$v'(x)=2x$$. Substituting into the rule gives

$$f'(x)=\dfrac{1\cdot(1+x^{2})-x\cdot 2x}{(1+x^{2})^{2}} =\dfrac{1+x^{2}-2x^{2}}{(1+x^{2})^{2}} =\dfrac{1-x^{2}}{(1+x^{2})^{2}}.$$

The numerator $$1-x^{2}=0$$ when $$x=\pm1$$, so the critical points are $$x=-1$$ and $$x=1$$. We evaluate $$f$$ there:

$$f(1)=\dfrac{1}{1+1^{2}}=\dfrac12, \qquad f(-1)=\dfrac{-1}{1+(-1)^{2}}=-\dfrac12.$$

Because the denominator $$\,(1+x^{2})^{2}\gt 0$$ for every real $$x$$, the sign of $$f'(x)$$ is the sign of $$1-x^{2}$$. Thus

$$ \begin{cases} f'(x)\gt 0 &\text{for }|x|\lt 1,\\[4pt] f'(x)\lt 0 &\text{for }|x|\gt 1. \end{cases} $$

So the function increases on $$(-1,1)$$, decreases on $$(-\infty,-1)$$ and on $$(1,\infty)$$. The point $$x=1$$ therefore gives a global maximum $$\dfrac12$$, and $$x=-1$$ gives a global minimum $$-\dfrac12$$. As $$x\rightarrow\pm\infty$$ we have

$$\lim_{x\to\pm\infty}\dfrac{x}{1+x^{2}} =\lim_{x\to\pm\infty}\dfrac{1}{x+1/x} =0,$$

so the graph approaches the $$x$$-axis from both sides without touching it.

Combining all this information, the set of output values is the open interval

$$\left(-\dfrac12,\dfrac12\right).$$

Because the codomain stated in the question is exactly this interval, every element of the codomain appears as an output. Hence the function is surjective (onto).

Next we test whether the function is one-one (injective). A continuous function that first decreases, then increases, and then decreases again cannot be monotonic on the whole of $$\mathbb R$$, so it is very likely not injective. We confirm this by finding two different inputs that give the same output.

Let us choose the value $$y=\dfrac25=0.4$$, which lies strictly between $$0$$ and $$\dfrac12$$. We solve the equation $$f(x)=y$$ algebraically.

Starting from $$\displaystyle y=\dfrac{x}{1+x^{2}}$$ we cross-multiply:

$$y(1+x^{2})=x \;\;\Longrightarrow\;\; y + yx^{2} - x = 0.$$

This is a quadratic in $$x$$:

$$y\,x^{2} - x + y = 0.$$

For a quadratic $$ax^{2}+bx+c=0$$ the solutions are given by the quadratic formula

$$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$

In our case $$a=y,\,b=-1,\,c=y$$, so

$$x=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4y^{2}}}{2y} =\dfrac{1\pm\sqrt{1-4y^{2}}}{2y}.$$

Because $$0\lt y\lt \dfrac12$$, the discriminant $$1-4y^{2}$$ is positive, giving two distinct real solutions. Putting our specific choice $$y=\dfrac25$$, we get

$$x_{1}=\dfrac{1-\sqrt{1-\dfrac{16}{25}}}{\dfrac{4}{5}} =\dfrac{1-\sqrt{\dfrac{9}{25}}}{\dfrac45} =\dfrac{1-\dfrac35}{\dfrac45} =\dfrac{\dfrac25}{\dfrac45} =\dfrac12,$$

and

$$x_{2}=\dfrac{1+\sqrt{1-\dfrac{16}{25}}}{\dfrac45} =\dfrac{1+\dfrac35}{\dfrac45} =\dfrac{\dfrac85}{\dfrac45} =2.$$

Thus $$f\!\left(\dfrac12\right)=\dfrac25=f(2).$$ Two different inputs map to the same output, so the function is not injective.

We have shown that

• $$f$$ is surjective onto its codomain $$\left(-\dfrac12,\dfrac12\right)$$,
• $$f$$ is not injective.

Therefore $$f$$ is surjective but not injective, which corresponds to Option C.

Hence, the correct answer is Option C.