[tex]f : \mathbb_{R} \rightarrow \mathbb_{R},$ $ \quad f(x) = x-\sqrt{x^2+1}\\ \\ f'(x) = 1 - \dfrac{(x^2+1)'}{2\sqrt{x^2+1}} \Rightarrow f'(x) = 1-\dfrac{2x}{2\sqrt{x^2+1} }\Rightarrow f'(x) = 1 - \dfrac{x}{\sqrt{x^2+1}}\\ \\ $Consideram $ g(x) = f'(x). \\ \\ $Monotonia se afla prin derivata functiei, derivam g(x):$ \\ \\ g'(x) = 0 - \dfrac{x'\cdot (\sqrt{x^2+1})-x\cdot(\sqrt{x^2+1})' }{x^2+1} \Rightarrow \\ \\ [/tex]
[tex]\Rightarrow g'(x) = -\dfrac{^{\Big{\sqrt{x^2+1}$ $\big)}}\sqrt{x^2+1}-x\cdot \dfrac{x}{\sqrt{x^2+1}}}{x^2+1} \Rightarrow \\ \\ \Rightarrow g'(x) = -\dfrac{\dfrac{x^2+1-x^2}{\sqrt{x^2+1}}}{x^2+1} \Rightarrow g'(x) = -\dfrac{x^2+1-x^2}{\sqrt{x^2+1}\cdot (x^2+1)}\Rightarrow \\ \\ \Rightarrow g'(x) = -\dfrac{1}{\sqrt{x^2+1}\cdot (x^2+1)}
\left\| \begin{array}{c}\sqrt{x^2+1}\geq0,\quad \forall x\in \mathbb_{R}$ $ \\ x^2+1}\geq0,\quad \forall x\in \mathbb_{R}$ $\end{array} \right | \Rightarrow
[/tex]
[tex]\Rightarrow g'(x)\ \textless \ 0,\quad \forall x\in \mathbb_{R} $ $ \Rightarrow g(x) \rightarrow $ strict descrescatoare pe $ \mathbb_{R}.[/tex]
