Answer
No inverse
Work Step by Step
Given the function
\[
f(x) \;=\; \frac{x}{x^2 + 1},
\quad \text{with domain } \mathbb{R}.
\]
We must check whether \(f\) is one-to-one (injective) and onto (surjective) \(\mathbb{R}\).
---
## 1. Is \(f\) One-to-One?
To test injectivity, suppose
\[
f(x_1) \;=\; f(x_2).
\]
That is,
\[
\frac{x_1}{x_1^2 + 1} \;=\; \frac{x_2}{x_2^2 + 1}.
\]
Cross-multiplying gives
\[
x_1(x_2^2 + 1) \;=\; x_2(x_1^2 + 1),
\]
which can be rearranged and factored into
\[
(x_2 - x_1)\,\bigl(x_1 x_2 - 1\bigr) \;=\; 0.
\]
Hence either \(x_2 = x_1\) (the usual case for injectivity) **or** \(x_1 x_2 = 1\).
- For example, if \(x_1 = 2\) and \(x_2 = \tfrac{1}{2}\), then \(x_1 x_2 = 1\) and indeed
\[
f(2) \;=\; \frac{2}{2^2 + 1} \;=\; \frac{2}{5},
\quad
f\!\Bigl(\tfrac12\Bigr) \;=\; \frac{\tfrac12}{\tfrac14 + 1} \;=\; \frac{\tfrac12}{\tfrac54}
\;=\; \frac{2}{5}.
\]
Yet \(2 \neq \tfrac12\).
Thus \(f\) is **not** injective: distinct inputs can yield the same output.
---
## 2. Is \(f\) Onto \(\mathbb{R}\)?
We ask if every real \(y\) has some \(x\) with \(f(x) = y\). That is,
\[
\frac{x}{x^2 + 1} \;=\; y
\quad\Longleftrightarrow\quad
x \;=\; y\,(x^2 + 1).
\]
Rewriting,
\[
y\,x^2 \;-\; x \;+\; y \;=\; 0.
\]
Viewed as a quadratic in \(x\) with coefficients \((y,\, -1,\, y)\), the discriminant is
\[
\Delta \;=\; (-1)^2 \;-\; 4\,y\,y \;=\; 1 \;-\; 4y^2.
\]
For \(x\) to be real, we need \(\Delta \ge 0\), i.e.
\[
1 \;-\; 4y^2 \;\ge\; 0
\quad\Longleftrightarrow\quad
4y^2 \;\le\; 1
\quad\Longleftrightarrow\quad
|y| \;\le\; \tfrac12.
\]
Hence \(f(x)\) can only take values in the **closed interval** \(\bigl[-\tfrac12,\, \tfrac12\bigr]\). It **never** produces values outside \(\pm \tfrac12\). Therefore \(f\) is **not** onto all of \(\mathbb{R}\); for instance, there is no \(x\) with \(f(x) = 1\).
---
## 3. Conclusion
Since \(f\) fails to be injective and also fails to be onto \(\mathbb{R}\), it is certainly **not** a one-to-one correspondence \(\mathbb{R} \to \mathbb{R}\). There is no inverse function when the codomain is taken to be all real numbers.
