Question

rezolvarea vreau, nu doar ideea.

Answer 1

$f(x) = x+\sqrt{x^2+1}\\ \\ a)\quad \lim\limits_{x\to 0}\dfrac{f(x)-1}{x} = \lim\limits_{x\to 0}\dfrac{x+\sqrt{x^2+1}-1}{x} \overset{\frac{0}{0}}{=} f'(0) \\ \\ f'(x) = 1+\dfrac{x}{\sqrt{x^2+1}} \Rightarrow f'(0) = 1\\ \\ \\ b)\quad \lim\limits_{x\to \infty} \Big(x+\sqrt{x^2+1}-mx-n\Big) = \\ \\ = \lim\limits_{x\to \infty}\Big[x+(x)-mx-n\Big] =\lim\limits_{x\to \infty}\Big[(2-m)x-n\Big] = 0$

$\Rightarrow m = 2,\quad n = 0\\ \\ \Rightarrow y = 2x\quad \text{asimptota oblica spre }+\infty\\ \\ \sqrt{x^2+1}\approx x,\quad x\to +\infty \\\\ \text{Fiindca }x^2 \approx (\sqrt{x^2+1})^2,\quad x\to+\infty$

$c)\quad f(x) = x+\sqrt{x^2+1} \\ \\ f'(x) = 1+\dfrac{x}{\sqrt{x^2+1}} = \dfrac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}} = 0 \\ \\ \Rightarrow x +\sqrt{x^2+1} = 0 \Rightarrow x = -\sqrt{x^2+1} \Rightarrow x^2 = x^2+1\Rightarrow x\in \emptyset\\ \\ f'(x) > 0 \Rightarrow f(x)\text{ strict crescatoare pe }\mathbb{R} \\ \\\text{Nu avem conditii de existenta} \Rightarrow f(x) \text{ continua pe }\mathbb{R}$

$\Rightarrow f(x) \text{ este bijectiva },\quad f(x) \in (0,+\infty) \\ \\ \Rightarrow f(x) = m \text{ are solutie unica in }\mathbb{R},\quad \forall m>0$