Question

Rezolvă expresia 100 pct​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

Am rezolvat subpunctul b.

$-2\leq E(x)\leq 2 \iff |E(x)|\leq 2 \ \forall x\in \mathbb{R}\setminus\{\pm1\}\\ |E(x)|=|\frac{4x}{x^2+1} |=\frac{|4x|}{x^2+1} =\frac{4|x|}{x^2+1} \\x^2+1 > 0 \ \forall x\in \mathbb{R}\setminus\{\pm1\}\\\frac{4|x|}{x^2+1} \leq 2\\4|x|\leq 2(x^2+1)\\2|x|\leq x^2+1$

$Cazul \ I:x \in (-\infty,0) \setminus\{-1\}\\-2x\leq x^2+1\\x^2+2x+1\geq 0\\(x+1)^2\geq 0 \ \forall x \in (-\infty,0) \setminus\{-1\}\\Cazul \ II:x\in[0,\infty)\setminus\{1\}\\2x\leq x^2+1\\x^2-2x+1\geq 0\\(x-1)^2\geq 0 \ \forall x \in [0,\infty)\setminus\{1\}$

$Deci \ |E(x)|\leq 2 \ \forall x \in \mathbb{R}\setminus\{\pm1\}$