How to find the range of $1 / (1+x^2)^{1/2}$? How to find the range of $$\frac{1}{\sqrt{1+x^2}}$$?
Ok. I've revised the (easy theory). I would like to complete the exercise finding the derivative of f(x) and setting equal to zero. I do it correctly, $$f'(x) = - x / (1+x^2)^{3/2}$$ but I don't have clear how to solve this equation once set equal to zero. I simplify the denominator but I don't know how to isolate the x.
 A: To find the range, since there is only one $x$ in the expression, you can simply proceed in steps:


*

*What is the range of $x^2$?

*What is, therefore, the range of $1+x^2$?

*What is, therefore, the range of $\sqrt{1+x^2}$?

*What is, therefore, the range of $\frac1{\sqrt{1+x^2}}$?


It may help to sketch each of these functions crudely as you proceed.
A: We try to rewrite $x$ in terms of $y$:
$$
0 < \frac{1}{\sqrt{1+x^2}} = y \le 1 \quad (x \in \mathbb{R})\Rightarrow \\
0 < \frac{1}{1+x^2} = y^2 \le 1 \Rightarrow \\
1 \le \frac{1}{y^2} = 1 + x^2 < \infty \Rightarrow \\
0 \le \frac{1-y^2}{y^2} = x^2 < \infty
$$
and get
$$
x = 
\pm \frac{\sqrt{1-y^2}}{y} \quad  (y \in (0,1])
$$
$y$ is not a function anymore, for every $y \in (0,1)$
there are two choices of $x$.
Here is an image:

Addendum: 
$$
0 \le x^2 \quad (x \in \mathbb{R}) \Rightarrow \\
1 \le 1 + x^2 \quad (x \in \mathbb{R}) \Rightarrow \\
\frac{1}{1+x^2} \le 1
$$
because $1+x^2 \ge 1$ thus $1+x^2 \ne 0$ we can divide both sides of the inequality by it, because $1+x^2 > 0$ the direction of the comparison operator did not change.
The RHS is $y^2$, so
$$
y^2 = \frac{1}{1+x^2} \le 1 \Rightarrow \\
y = \frac{1}{\sqrt{1+x^2}} \le \sqrt{1} = 1
$$
This one holds because $\sqrt{.}$ is monotone: 
$$
x_1 \le x_2 \Rightarrow \sqrt{x_1} \le \sqrt{x_2}
$$
Addendum:
A criterion for local extrema is the vanishing of the first derivative:
$$
y = \frac{1}{\sqrt{1+x^2}} \Rightarrow \\
y' = -\frac{1}{2\left(1+x^2\right)^{3/2}}(2x) = -\frac{x}{\left(1+x^2\right)^{3/2}}
$$
This means
$$
0 = y' = -\frac{x}{\left(1+x^2\right)^{3/2}} \Rightarrow x = 0
$$
Thus $y(0) = 1$ might be a local extremum. To know more we look at
the second derivative:
$$
y'' 
= 
-\frac{\left(1+x^2\right)^{3/2} 
- x \frac{3}{2}\left(1+x^2\right)^{1/2}(2x)}
{\left(1+x^2\right)^3}
=
-\frac{\left(1 - 2 x^2 \right) \left(1 + x^2\right)^{1/2}}
{\left(1+x^2\right)^3}
= \frac{2x^2 - 1}
{\left(1+x^2\right)^{5/2}}
$$
We see $y''(0) = -1 \ne$ so it is a local maximum. 

