Let $$f : R \to R$$ be defined by $$f(x) = \frac{x}{1+x^2}$$, $$x \in R$$. Then the range of $$f$$ is

We have a real-valued function $$f:\mathbb R \to \mathbb R$$ defined by

$$f(x)=\dfrac{x}{1+x^{2}}, \qquad x\in\mathbb R.$$

To determine the range of $$f$$ we need all possible values taken by $$\dfrac{x}{1+x^{2}}$$ as $$x$$ varies over every real number.

Because the expression is a quotient of differentiable functions and the denominator $$1+x^{2}$$ is never zero, $$f(x)$$ is continuous for every real $$x$$. For a continuous function on the entire real line, extreme values (if any) will occur either at critical points (where the derivative is zero or undefined) or as $$x\to\pm\infty$$.

First we compute the derivative. We state the Quotient Rule: if $$g(x)=\dfrac{u(x)}{v(x)}$$ with both $$u, v$$ differentiable and $$v(x)\neq0$$, then

$$g'(x)=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^{2}}.$$

Here $$u(x)=x$$ and $$v(x)=1+x^{2}$$. Their derivatives are $$u'(x)=1$$ and $$v'(x)=2x$$. Substituting into the Quotient Rule we obtain

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

A critical point occurs where $$f'(x)=0$$ or $$f'(x)$$ is undefined. The denominator $$\bigl(1+x^{2}\bigr)^{2}$$ is always positive, so the only way $$f'(x)=0$$ is when the numerator $$1-x^{2}=0$$. Solving,

$$1-x^{2}=0 \;\Longrightarrow\; x^{2}=1 \;\Longrightarrow\; x=\pm1.$$

Thus, the critical points are $$x=1$$ and $$x=-1$$. We now evaluate $$f(x)$$ at these points:

For $$x=1$$:

$$f(1)=\dfrac{1}{1+1^{2}}=\dfrac{1}{2}.$$

For $$x=-1$$:

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

Next, we study the behaviour of $$f(x)$$ as $$x$$ tends to infinity or minus infinity. We write

$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\dfrac{x}{1+x^{2}} =\lim_{x\to\infty}\dfrac{1/x}{1/x^{2}+1} =\dfrac{0}{0+1}=0.$$

Similarly,

$$\lim_{x\to-\infty}f(x)=0.$$

So the function approaches $$0$$ from both sides but never exceeds (in magnitude) the values found at the critical points. To confirm that $$\dfrac{1}{2}$$ is the greatest and $$-\dfrac{1}{2}$$ is the least value, we examine the sign of the derivative.

Using $$f'(x)=\dfrac{1-x^{2}}{(1+x^{2})^{2}}$$, notice

• When $$|x|<1$$, the numerator $$1-x^{2}>0$$, so $$f'(x)>0$$ and the function is increasing.
• When $$|x|>1$$, the numerator $$1-x^{2}<0$$, so $$f'(x)<0$$ and the function is decreasing.

Hence $$x=1$$ gives a local maximum value $$\dfrac{1}{2}$$, and $$x=-1$$ gives a local minimum value $$-\dfrac{1}{2}$$. There are no larger or smaller values because the function decreases as we move away from these points and approaches $$0$$ asymptotically.

Therefore the set of all values attained by $$f(x)$$ is the closed interval

$$\left[-\dfrac{1}{2},\,\dfrac{1}{2}\right].$$

So the range is exactly $$\left[-\dfrac{1}{2},\,\dfrac{1}{2}\right]$$, which corresponds to Option A.

Hence, the correct answer is Option A.