The function f : R → [–1/2, 1/2] defined as f(x) = x / (1 + x²), is: (1) invertible. (2) injective b

Q: Wait, why is the function not injective? I thought every $x$ had a unique $y$?

That is the definition of a function, but injectivity is stricter! It requires that no two different $x$ values share the same $y$. As we saw, $f(2) = 2/5$ and $f(1/2) = 2/5$. Since $2 \neq 1/2$, the function is many-to-one, not one-to-one.

Q: Why do we use the discriminant $D \geq 0$ to find the range?

Because $y = \frac{x}{1 + x^2}$ implies $yx^2 - x + y = 0$. For any $y$ in the range, there must exist at least one real $x$ that satisfies this equation. The condition for a quadratic to have real roots is $D \geq 0$. This is a powerful algebraic tool to map the output space!

Q: Can we use calculus to find the range instead of the discriminant?

Absolutely! You could find the critical points by setting $f'(x) = 0$, evaluate the function at those points, and check the limits as $x \to \pm \infty$. Both methods are elegant, but the discriminant method is often faster for rational functions like this one.

Question

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

Options:

(A) invertible.

(B) injective but not surjective.

(C) surjective but not injective.

(D) neither injective nor surjective.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of mathematics! Today, we are going to peel back the layers of a seemingly simple function:

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

This function is defined from the set of all real numbers $\mathbb{R}$ to the codomain $[-\frac{1}{2}, \frac{1}{2}]$. Let us embark on this journey to determine if it is injective (one-to-one) or surjective (onto).

### The Injectivity Trap

Imagine you are a detective. Your job is to determine if every output $y$ has exactly one unique input $x$ that created it. If you find even one case where two different inputs $x_1$ and $x_2$ lead to the same output $y$, the function is not injective.

Let us test this. Suppose we pick $x = 2$:

$$f(2) = \frac{2}{1 + 2^2} = \frac{2}{5}$$

Now, what if we pick $x = 1/2$?

$$f(1/2) = \frac{1/2}{1 + (1/2)^2} = \frac{1/2}{5/4} = \frac{2}{5}$$

Look at that! Since $f(2) = f(1/2)$ but $2 
eq 1/2$, this is the smoking gun. Visually, a horizontal line drawn at $y = 2/5$ would slice through the graph at two distinct points, confirming that our function is **not injective**.

### The Algebraic Dance for Surjectivity

A function is surjective if its range covers every single value in its codomain. We are given the codomain $[-\frac{1}{2}, \frac{1}{2}]$. To see if the function hits every value in this interval, we set $y = \frac{x}{1 + x^2}$ and rearrange it into a quadratic equation:

$$y(1 + x^2) = x \Rightarrow yx^2 - x + y = 0$$

For $x$ to be a real number, the discriminant $D$ of this quadratic must be non-negative. Recall that for $ax^2 + bx + c = 0$, the discriminant is $D = b^2 - 4ac$. Here, $a = y$, $b = -1$, and $c = y$:

$$D = (-1)^2 - 4(y)(y) = 1 - 4y^2$$

For real roots, we require $1 - 4y^2 \geq 0$. This simplifies to $4y^2 \leq 1$, or $y^2 \leq 1/4$. Taking the square root, we get $|y| \leq 1/2$, which means $y \in [-\frac{1}{2}, \frac{1}{2}]$.

### The Final Verdict

We have calculated the range to be $[-\frac{1}{2}, \frac{1}{2}]$. Since this calculated range is exactly equal to the codomain provided in the problem, the function is **surjective**.

We have successfully navigated the traps of injectivity and the algebraic rigor of surjectivity. The function is **surjective but not injective**. Keep exploring, keep questioning, and remember that every equation is just a story waiting to be told!

Animated Solution for Mathematics - Functions: The function $f : R \to [- \frac{1}{2}, \frac{1}{2}]$ defined as $f (x) = \frac{x}{1 + x ^{2}}$ , is:

Select Answer:

Visualized Solution

Function Definition and Codomain

Checking Injectivity (One-to-One)

Counterexample for Injectivity

The Horizontal Line Test

Analyzing Surjectivity (Onto)

Setting up for Range Calculation

Applying the Discriminant Condition

Solving for the Range

Final Verdict

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Injectivity Trap

The Algebraic Dance for Surjectivity

The Final Verdict

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