Given a problem as follows.
Find the range of $$ f(x)=\frac{4x}{\sqrt{(25+x^2)(1+x^2)}} $$ for any real $x$.
My Attempt
\begin{gather} y=\frac{4x}{\sqrt{(25+x^2)(1+x^2)}}\\ \sqrt{(25+x^2)(1+x^2)}=\frac{4x}{y}\\ x^4+26x^2+25 = \frac{16}{y^2}x^2\\ x^4+(26-\frac{16}{y^2})x^2+25=0 \end{gather}
As $y$ is defined for any real $x$ then $x^2$ must be non-negative real numbers.
The constraints are \begin{gather} D\geq 0 \quad\text{and}\quad -\frac{b}{a} \geq 0 \quad\text{and}\quad \frac{c}{a} \geq 0 \end{gather}
As you can see, the third constraint has been satisfied so only the first two constraints will be considered as follows.
First constraint: \begin{gather} (26-\frac{16}{y^2})^2 - 4\times 1\times 25 \geq 0\\ \frac{64(y-1)(y+1)(3y-2)(3y+2)}{y^4}\geq 0 \end{gather}
The solution is $(-\infty,-1]\cup[-2/3,0)\cup(0,2/3]\cup[1,\infty)$.
Second constraint: \begin{gather} -\frac{26-\frac{16}{y^2}}{1} \geq 0\\ 26-\frac{16}{y^2} \leq 0 \\ \frac{(y\sqrt{26}-4)(y\sqrt{26}+4)}{y^2} \leq 0 \end{gather}
The solution is $[-4/\sqrt{26},0)\cup(0,4/\sqrt{26}]$.
By intersecting the above solutions, the range of $f(x)$ is $[-2/3,0)\cup(0,2/3]$.
Question
If we consider the $f(x)$, it will be zero if $x=0$. However the previously calculated range says it never reach zero. How to avoid this contradictory result?
When $x = 0$, we have $f(x) = 0$ from the definition of $x$, which you would write as $y = 0$.
In the second line of your "attempt", you divide by $y$, but fail to note that this is only valid (as are all subsequent steps) if $y \ne 0$ (hence $x \ne 0$).
So the answer is: keep track of your domains, especially when dividing by anything or when squaring (or applying other non-injective functions) both sides of an equality.