If we know range of a function , how can construct range of other function?

Question

If we know $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$ how can obtain the range of the function below? $$y=\frac{4 x}{9x^2+25}$$

The problem is what's the range with respect to $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$

Answer 1

Write $$\frac{4x}{9x^2+25}=\frac{4}{9x+\frac{25}{x}}$$ and use AM-GM in two cases:

1) $x>0$: $$y\leq\frac{4}{2\sqrt{9x\cdot\frac{25}{x}}}=\frac{2}{15}.$$

2) $x<0$: $$y\geq-\frac{4}{2\sqrt{(-9x)\cdot\left(-\frac{25}{x}\right)}}=-\frac{2}{15}.$$ The equality occurs for $9x=\frac{25}{x}$ and since we have a continuous function,

we got the answer: $$\left[-\frac{2}{15},\frac{2}{15}\right]$$

Answer 2

$$y=\frac{4 x}{9x^2+25}\\y=\frac{4 x}{25(\frac{9}{25}x^2+1)}=\\y=\frac4{25}\frac{ x}{((\frac{3}{5}x)^2+1)}$$ now take $u=\frac{3x}{5}$ $$y=\frac4{25}\frac{ x}{((\frac{3}{5}x)^2+1)}=\\\frac4{25}\frac{ \frac{5}{3}u}{((u)^2+1)}=\\\frac{4}{15}\frac{u}{u^2+1}\\$$you know $\frac{-1}{2}\leq \frac{u}{u^2+1}\leq \frac{1}{2}$ so $$-\frac{4}{15}\times \frac {1}{2} \leq y=\frac{4}{15}\frac{u}{u^2+1} \leq \frac{4}{15}\times \frac {1}{2}$$

Answer 3

We have \begin{eqnarray*} - \frac{1}{2} \leq \frac{x}{x^2+1} \leq \frac{1}{2} \end{eqnarray*} Let $x= \frac{3z}{5}$ \begin{eqnarray*} - \frac{1}{2} \leq \frac{\frac{3z}{5}}{(\frac{3z}{5})^2+1} \leq \frac{1}{2} \\ - \frac{1}{2} \leq \frac{15z}{9z^2+25} \leq \frac{1}{2} \end{eqnarray*} Now multiply by $ \frac{4}{15}$ \begin{eqnarray*} - \frac{2}{15} \leq \frac{4z}{9z^2+25} \leq \frac{2}{15}. \end{eqnarray*}