Show that $[a,b] \subset Range(f)$

Question

Question. Show that $[-\frac{1}{2},\frac{1}{2}]\subset Range(f)$, where $f(x)=\frac{x}{x^2+1}$.

My proof.

• Let $y=\frac{x}{x^2+1}\Leftrightarrow yx^2-x+y=0.$

• For real values of y, values of x must be real:

$Discriminant\geq0\Leftrightarrow 1-4y^2\geq0\Leftrightarrow -\frac{1}{2}\leq y\leq\frac{1}{2}$.

• Therefore, $[-\frac{1}{2},\frac{1}{2}]\subset Range(f)$.

Are there other solutions to this type of questions?

Answer 1

Yes. Actually you proved that $[-\frac{1}{2},\frac{1}{2}]=\text{Range}(f)$.

To prove the inclusion you want with less work, just notice that $f(1)=\frac{1}{2}, f(-1)=-\frac{1}{2}$ and conclude by the intermediate value theorem.

Answer 2

• $f$ is continuous, so if $a,b\in\mathrm{Range}(f)$ then $[a,b]\subseteq\mathrm{Range}(f)$ by the IVT.

• $f$ is odd, so if $a\in\mathrm{Range}(f)$ then $-a\in\mathrm{Range}(f)$.

• $f(1)=\frac{1}{2}$, so $\frac{1}{2}\in\mathrm{Range}(f)$.

Note that the second bullet is overkill here, since you can just show directly that $-\frac{1}{2}\in\mathrm{Range}(f)$ by noting that $f(-1)=-\frac{1}{2}$.