Given function

$f:\mathbb{R}\to \left[-\frac12,\frac12\right], \qquad f(x)=\frac{x}{1+x^2}$

We must check whether it is injective and/or surjective onto the codomain $\left[-\frac12,\frac12\right]$.

Check injectivity

A function is injective if different inputs always give different outputs.

Let us test simple values:

$f(2)=\frac{2}{1+4}=\frac25$ $f\left(\frac12\right)=\frac{\frac12}{1+\frac14}=\frac{\frac12}{\frac54}=\frac{2}{5}$

Since $2\neq \frac12 \quad \text{but} \quad f(2)=f\left(\frac12\right),$ the function is not injective.

Check surjectivity onto $\left[-\frac12,\frac12\right]$

We need to show that for every $y\in\left[-\frac12,\frac12\right],$ there exists some $x\in\mathbb R$ such that $\frac{x}{1+x^2}=y.$

Solve for $x$:

$y=\frac{x}{1+x^2}$ $y(1+x^2)=x$ $yx^2-x+y=0$

This is a quadratic in $x$. For real $x$ to exist, its discriminant must be non-negative:

$\Delta = (-1)^2-4(y)(y)=1-4y^2$

Now if $y\in\left[-\frac12,\frac12\right],$ then $1-4y^2\ge 0.$

Hence the quadratic has a real solution for every such $y$. Therefore for every $y\in\left[-\frac12,\frac12\right]$, there exists real $x$ with $f(x)=y$.

So $f$ is surjective onto $\left[-\frac12,\frac12\right]$.

Conclusion

• Not injective
• Surjective

Therefore the correct option is $\boxed{\text{C: surjective but not injective}}$

Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.