Question

haideti va rog frumos.​

Answer 1

$f(x) = \dfrac{x^2+1 +2x-2}{x^2+1} = 1 +2\cdot \dfrac{x-1}{x^2+1}\\ \\ f'(x) = \dfrac{x^2+1-2x(x-1)}{(x^2+1)^2} = \dfrac{-x^2+2x+1}{(x^2+1)^2}\\ \\ f'(x) = 0 \Rightarrow -x^2 +2x+1 = 0 \\ \\ \Delta = 8\Rightarrow x_{1,2} = \dfrac{-2\pm 2\sqrt 2}{-2} =\pm \sqrt 2 +1$

$\lim\limits_{x\to\pm \infty}f(x) = 1,\quad f(-\sqrt 2+1) = \dfrac{2-2\sqrt 2+1-2\sqrt 2+2-1}{4-2\sqrt 2}=\\ \\ =\dfrac{4-4\sqrt 2}{4-2\sqrt 2} = \dfrac{2-2\sqrt 2}{2-\sqrt 2} = \dfrac{4+2\sqrt 2 - 4\sqrt 2 -4}{4-2} = -\sqrt 2\\ \\\\ \Rightarrow f_{min} = -\sqrt 2 \Rightarrow \boxed{a\in\Big(-\infty, -\sqrt 2\Big]}$

Answer 2

Răspuns:

a€(-oo; -rad2]

Explicație pas cu pas:

In poza