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.
On
To find the range, since there is only one $x$ in the expression, you can simply proceed in steps:
It may help to sketch each of these functions crudely as you proceed.
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.