Find the domain and range of the function f(x) = x² / (1 + x²). Is the function one-to-one?

Q: Wait, why is the domain all real numbers? Can't the denominator be zero?

That is a fantastic question! It shows you are thinking about the 'danger zones' of rational functions. In a rational function, the denominator cannot be zero. Here, our denominator is $1 + x^2$. For this to be zero, we would need $x^2 = -1$. But in the set of real numbers, the square of any number is always greater than or equal to zero. Thus, $x^2$ can never be $-1$, and $1 + x^2$ will always be at least $1$. Since it is never zero, the function is safe for all real inputs!

Q: Why does the inequality sign flip when we take the reciprocal?

Think about it with numbers. We know that $2 0.25$, the inequality sign must flip! The same logic applies to our function. Since $1 + x^2 \geq 1$, its reciprocal $\frac{1}{1 + x^2}$ must be $\leq 1$. It is a fundamental property of inequalities that you must always watch out for!

Q: Why is the function not one-to-one? Can't we just solve for x?

You can try to solve for $x$, but you will run into a problem. If you set $f(x) = y$, you get $y = \frac{x^2}{1 + x^2}$, which leads to $y(1 + x^2) = x^2$, or $y + yx^2 = x^2$, which simplifies to $y = x^2(1 - y)$, so $x^2 = \frac{y}{1 - y}$. When you take the square root, you get $x = \pm \sqrt{\frac{y}{1 - y}}$. The $\pm$ sign is the smoking gun! It means for every positive $y$, there are two possible $x$ values (a positive and a negative one). That is exactly why it fails the one-to-one test.

Question

Find the domain and range of the function $f(x) = \frac{x^2}{1 + x^2}$. Is the function one-to-one?

Sigma Percentile · Accepted Answer

### The Dance of the Rational Function: A Journey into Domain and Range

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are dissecting the soul of a function. We are looking at the function defined by:

$$f(x) = \frac{x^2}{1 + x^2}$$

At first glance, it looks like a simple ratio of two parabolas, but beneath this surface lies a beautiful structure that reveals much about the nature of functions.

### Phase 1: The Domain - Setting the Stage

Imagine you are an architect designing a bridge. You need to know where the ground is solid and where it is unstable. In the world of functions, the domain is our solid ground, defined as the set of all inputs $x$ for which the function is defined.

Our function is a rational one, a fraction. The golden rule of fractions is simple: the denominator must never be zero. If it were, the function would explode into infinity, a singularity we must avoid.

We look at our denominator: $1 + x^2$. We ask ourselves, "Can this ever be zero?" We know that for any real number $x$, $x^2 \geq 0$.

If we add $1$ to both sides, we get $1 + x^2 \geq 1$. This means the denominator is always at least $1$ and is strictly positive. It never touches zero; therefore, there are no "forbidden" values for $x$. The domain is the entire real line, $\mathbb{R}$. The bridge is solid everywhere.

### Phase 2: The Range - The Art of Manipulation

Now that we know where we can stand, we want to know how high we can reach. The range is the set of all possible outputs $f(x)$.

We have $f(x) = \frac{x^2}{1 + x^2}$. To find the range, we use the elegance of algebra to simplify the expression so that $x$ appears only once. This is a classic JEE strategy: manipulate the numerator to look like the denominator.

We add $1$ and subtract $1$ in the numerator:

$$f(x) = \frac{x^2 + 1 - 1}{1 + x^2}$$

Now, we split the fraction:

$$f(x) = \frac{x^2 + 1}{1 + x^2} - \frac{1}{1 + x^2} = 1 - \frac{1}{1 + x^2}$$

This is a revelation! We have transformed a complex fraction into a simple subtraction. Now, we can build the range step-by-step.

We know $x^2 \geq 0$, which implies $1 + x^2 \geq 1$. Taking the reciprocal, we get:

$$0 < \frac{1}{1 + x^2} \leq 1$$

Now, multiply by $-1$ to reverse the inequalities:

$$-1 \leq -\frac{1}{1 + x^2} < 0$$

Finally, add $1$ to all parts:

$$1 - 1 \leq 1 - \frac{1}{1 + x^2} < 1 - 0$$

This simplifies to $0 \leq f(x) < 1$. The range is the interval $[0, 1)$. The function starts at $0$ and climbs, but it never quite reaches $1$. It approaches $1$ as $x$ goes to infinity, creating a horizontal asymptote at $y = 1$.

### Phase 3: The One-to-One Test - Symmetry and Mirrors

Finally, we ask: is this function one-to-one? A function is one-to-one if every output comes from a unique input. If $f(x_1) = f(x_2)$, then $x_1$ must equal $x_2$.

Let's test this. If we plug in $x = 1$, we get:

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

If we plug in $x = -1$, we get:

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

We have two different inputs, $1$ and $-1$, producing the same output, $\frac{1}{2}$. This is the definition of a function that is **NOT one-to-one**.

Graphically, this is because the function is symmetric about the $y$-axis—it is an even function. If you draw a horizontal line at $y = 0.5$, it hits the graph at two points, failing the horizontal line test.

Keep practicing these manipulations. Every time you simplify a fraction or bound an inequality, you are not just doing math; you are learning to see the hidden structure of the universe. Keep going!

Animated Solution for Mathematics - Functions: Find the domain and range of the function $f (x) = \frac{x ^{2}}{1 + x ^{2}}$ . Is the function one-to-one?

Visualized Solution

Visualizing the Function

Condition for Domain

Evaluating the Denominator

Concluding the Domain

Strategy for Range

Algebraic Manipulation

Splitting the Fraction

Bounding the Denominator

Taking the Reciprocal

Finalizing the Bounds

Concluding the Range

Checking One-to-One Nature

Counterexample

Final Conclusion

The Sigma Insight: Domain and Range of a Function

The Dance of the Rational Function

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the domain and range of the function f(x)=1+x2x2​. Is the function one-to-one?