reciprocal factor of absolute value when evaluating a square root expression Learning with an old russian math book, i found the following evaluation for the function $f(x)=\sqrt{1+x^2}$:
$f(\frac1x)=\vert x \vert^{-1}\sqrt{1+x^2}$
My evaluation gave me 
$\sqrt{1+\dfrac1{x^2}}$
Are these two statements algebraically equivalent? I don't see the connection to the absolute values here.
Any help is appreciated.
 A: As $x^2+1>x^2\ge0$
$$\sqrt{1+\dfrac1{x^2}}=\dfrac{\sqrt{x^2+1}}{\sqrt{x^2}}$$
Now for real $x,$  $$\sqrt{x^2}=|x|=\begin{cases} x &\mbox{if } x\ge0 \\ 
-x & \mbox{if } x<0\end{cases}$$
A: Ah, good old Russian math books. :)
If there's one thing the USSR did right, it's STEM education.

I assume that the following is really what you care about. 

Why $\lvert x \rvert$ and not $x$?
The $x^2$ in $\sqrt{x^2}$ will make the contents of the root positive,
  regardless if $x$ is positive or negative. Taking the root of the
  positive number will result in another positive value.

In case you need more explanation, continue reading.

We start with this function:
$$ f(x) = \sqrt{1 + x^2} $$

Now, input $\frac{1}{x}$ as the $x$ parameter.
If you're not sure why $\frac{1}{x}$ is equal to $x$, well it's not. The $x$ in the reciprocal is not the same $x$ as the function parameter.
It might help you to see $ f(a) = \sqrt{1 + a^2} $ where $ a = \frac{1}{x} $ in such specific case.

$$ f(\frac{1}{x}) = \sqrt{1 + (\frac{1}{x})^2}$$
$$ = \sqrt{1 + \frac{1}{x^2}} $$
$$ = \sqrt{\frac{x^2 + 1}{x^2}} $$
$$ = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2}} $$
$$ = \frac{\sqrt{x^2 + 1}}{\lvert x \rvert} $$
$$ = \lvert x \rvert^{-1}\sqrt{x^2 + 1} $$
A: Notice that we can write $f(x)$ as:
$$ f(x) = \sqrt{x^2\left(\frac{1}{x^2} + 1\right)}=|x| \cdot \sqrt{\left(\frac{1}{x^2} + 1\right)}.$$
Thus, 
$$f\left(\frac 1x\right)=\left|\frac 1x\right| \cdot \sqrt{\frac{1}{\frac{1}{x^2}} +1  }= \frac {1}{|x|}\cdot \sqrt{x^2+1}$$
Also, your approach is right.
