[tex]\it\ a)\ f(x) = \dfrac{x}{x+1}-ln(x+1)
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f'(x) = \left(\dfrac{x}{x+1}\right)'-(ln(x+1))'=\dfrac{x+1-x}{(x+1)^2}-\dfrac{1}{x+1} =\\ \\ \\
=\dfrac{1}{(x+1)^2} - \dfrac{^{x+1)}1}{\ \ \ x+1} = \dfrac{1-x-1}{(x+1)^2}=-\dfrac{x}{(x+1)^2}[/tex]
[tex]\it\ b)\ f'(x) = -\dfrac{x}{(x+1)^2}\ \ \ \ (1)
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x\in[0,\ \ \infty) \Rightarrow x\geq 0|_{\cdot(-1)} \Rightarrow -x\leq 0 \ \ \ (2)
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(1),\ (2) \Rightarrow f'(x) <0 ,\ \forall\ x\in(0,\ \ \infty) \Rightarrow f(x) \ strict\ descresc\breve{a}toare[/tex]
[tex]\it\ c)\ f(0) = 0-ln1=0-0=0\ \ \ \ (3)
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f(x) \ strict\ descresc\breve{a}toare \ \ \ \ \ \ (4)
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(3),\ (4) \Rightarrow f(x) \leq0 \Rightarrow \dfrac{x}{x+1}-ln(x+1) \leq0 \Rightarrow
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\Rightarrow \dfrac{x}{x+1} \leq ln(x+1), \ \ \forall x\in [0,\ \ \infty)
[/tex]
