Question

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Answer 1

Răspuns:

$a \in [1, +\infty)$

Explicație pas cu pas:

$f' (x)=a-\frac{2x}{1+x^2}\\ f\: crescatoare \iff f'(x) \geq 0, (\forall) x \in \mathbb{R}\\ \implies a \geq \frac{2x}{1+x^2}, (\forall) x \in \mathbb{R}$

$Fie\: g :\mathbb{R} \to \mathbb{R}, g(x) = \frac{2x}{1+x^2}.\\ \\ \textrm{Aflam maximul acestei functii}.\\ g' = 0 \iff \frac{1+x^2-x\cdot2x}{{(1+x^2)}^2} = 0 \iff 1 - x^2 = 0 \iff x = \pm 1\\ g(1) = 2 \cdot \frac{1}{2} = 1\\ g(-1) = \frac{-2}{2} = -1 \\ \ \implies \textrm{Maximul functiei g este 1}$

$\implies a \geq 1 \iff a \in [1, +\infty)$