Question

La 1 punctul c).........
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Answer 1

$f:\mathbb{R}\to \mathbb{R},\, f(x) = \dfrac{x^2}{x^2+4}\\ \\ f'(x) = \dfrac{2x(x^2+4)-2x(x^2)}{(x^2+4)^2} = \dfrac{2x(x^2+4-x^2)}{(x^2+4)^2} = \dfrac{8x}{(x^2+4)^2}\\ \\ f'(x) = 0 \Rightarrow 8x = 0 \Rightarrow x = 0\\ \\ \text{Se observa ca:} \\ \\f'(x) < 0,\quad x <0 \\ f'(x) > 0,\quad x>0\\ \\ \lim\limits_{x\to -\infty}f(x) = 1\\ \lim\limits_{x\to+\infty}f(x) = 1\\ \\ \text{Din tabelul de variatie din imagine rezulta ca:} \quad 0\leq f(x) <1,\quad \forall x\in \mathbb{R}$