Question

f:R---->R, f(x)= x - radical din(x^2 +1)
Aratati ca derivata functiei este descrescatoare pe R

Answer 1

$f : \mathbb_{R} \rightarrow \mathbb_{R}, \quad f(x) = x-\sqrt{x^2+1}\\ \\ f'(x) = 1 - \dfrac{(x^2+1)'}{2\sqrt{x^2+1}} \Rightarrow f'(x) = 1-\dfrac{2x}{2\sqrt{x^2+1} }\Rightarrow f'(x) = 1 - \dfrac{x}{\sqrt{x^2+1}}\\ \\ Consideram g(x) = f'(x). \\ \\ Monotonia se afla prin derivata functiei, derivam g(x): \\ \\ g'(x) = 0 - \dfrac{x'\cdot (\sqrt{x^2+1})-x\cdot(\sqrt{x^2+1})' }{x^2+1} \Rightarrow$


[tex]\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]

$\Rightarrow g'(x)\ \textless \ 0,\quad \forall x\in \mathbb_{R} \Rightarrow g(x) \rightarrow strict descrescatoare pe \mathbb_{R}.$