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

Welcome, future engineers! Today, we are going to peel back the layers of a function that might look simple but holds a beautiful geometric secret. We are looking at the function:

When you see a rational function like this, don't just stare at it—interrogate it! Ask yourself: "Where does this curve go?" and "Does it climb to infinity, or does it get trapped in a box?"

The range is the set of all possible outputs, the "shadow" this function casts on the $y$-axis. To find this, we need to identify the peaks and valleys using calculus as our flashlight in the dark.

To find the range, we need to know how high and how low this graph goes. In other words, we need its maximum and minimum values. Our best tool for finding extreme values is the derivative.

Since our function is a fraction, we must use the Quotient Rule. Remember, the derivative of $\frac{u}{v}$ is given by:

Our numerator $u$ is $x$, so its derivative $u^{\prime }$ is $1$. Our denominator $v$ is $1+x^2$, making its derivative $v^{\prime }$ equal to $2x$. Substituting these into our formula, we get:

Now, let's simplify that numerator. We have $1+x^2$, and then we subtract $x(2x)$, which is $2x^2$.

So, $1+x^2-2x^2$ simplifies beautifully to just $1-x^2$. Our denominator remains the same, yielding the simplified derivative:

This derivative is the key to the entire problem. It tells us exactly where the slope of the function is zero—the points where the curve stops climbing and starts falling, or vice versa.

To find the critical points, we set $f^{\prime }(x)=0$. For a fraction to be zero, its numerator must be zero.

Setting $1-x^2=0$ gives us $x^2=1$, which results in critical points at $x=1$ and $x=-1$.

At $x=1$, we evaluate the function:

This is our maximum peak. At $x=-1$, we evaluate:

This is our minimum valley.

We must also check the behavior of the function at the extreme ends of the $x$-axis, as $x$ approaches positive and negative infinity.

As $x$ gets infinitely large, the $x^2$ term in the denominator grows much faster than the $x$ in the numerator. This means the fraction gets smaller and smaller, approaching $0$.

The same happens as $x$ approaches negative infinity. The curve flattens out towards the $x$-axis.

Visualize this: the function rises from zero, hits a maximum height of $\frac{1}{2}$ at $x=1$, and then drops back towards zero. On the negative side, it dips to a minimum depth of $-\frac{1}{2}$ at $x=-1$, and then rises back to zero.

Since the function is continuous everywhere, it must take all $y$-values between its absolute minimum and absolute maximum.

Therefore, the complete range of the function $f(x)$ is the closed interval:

This is the elegance of calculus—combining algebraic manipulation with graphical intuition to solve the problem with absolute certainty. Keep practicing, and you will master these concepts!