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\begin{document}%%%%%%%%%%b{gedeonInfo}%%%%%%%%%
%%--Document N° 1508219666010--
%%--Corrigé du document N° 1508219405941--
%%%%%%%%%%e{gedeonInfo}%%%%%%%%%



\begin{claim}
\begin{enumerate}


\item Faux, on ne sait pas si :  $x\in[-1\,;\,+\infty[\Rightarrow f(x)\geq0$. 

\item
\begin{enumerate}


\item Vrai, car : $x\in\mathcal{S}\Leftrightarrow f(x)\geq0\Rightarrow
x\geq-1\Leftrightarrow x\in[-1\,;\,+\infty[$. 

\item Faux, si $x\geq-1$ on ne peut rien conclure sur le signe de  $f(x)$ (pas
de symbole ``$\Leftarrow$'') 
\end{enumerate}

\item Vrai car si $x\in]-\infty\,;\,-1[$ alors il est impossible que
$f(x)\geq0$ (sinon on aurait $x\geq-1$), donc $f(x)<0$. 

\item Faux, par exemple, en prenant $f(x)=1-x^{2}$ on a :\newline%
$f(x)\geq0\Leftrightarrow x\in[-1\,;\,1]\Rightarrow x\geq-1$ mais
$f(x)<0\nRightarrow x\in]-\infty\,;\,-1[$ car $f(2)<0$. 

\item Vrai puisque la fonction $f$ est définie sur $\mathbb{R}$. 
\end{enumerate}
\end{claim}


\end{document}