Bonjour Ueueuru
[tex]f(x)=\sqrt{-x^2+2x+1}[/tex]
1) Ensemble de définition.
[tex]\text{Condition : }-x^2+2x+1\ge0\\\\\text{Racines : }-x^2+2x+1=0\\\\\Delta=2^2-4\times(-1)\times1=4+4=8\ \textgreater \ 0\\\\x_1=\dfrac{-2-\sqrt{8}}{2\times(-1)}=\dfrac{-2-2\sqrt{2}}{-2}=\dfrac{-2(1+\sqrt{2})}{-2}=1+\sqrt{2}\\\\x_2=\dfrac{-2+\sqrt{8}}{2\times(-1)}=\dfrac{-2+2\sqrt{2}}{-2}=\dfrac{-2(1-\sqrt{2})}{-2}=1-\sqrt{2}\\\\\\\begin{array}{|c|ccccccc|} x&-\infty&&1-\sqrt{2}&&1+\sqrt{2}&&+\infty\\-x^2+2x+1&&-&0&+&0&-&\\ \end{array}\\\\\\\boxed{-x^2+2x+1\ge0\Longleftrightarrow x\in[1-\sqrt{2};1+\sqrt{2}]}[/tex]
Par conséquent,
l'ensemble de définition de f est [tex]\boxed{D_f=[1-\sqrt{2};1+\sqrt{2}]}[/tex]
2) Tableau de variation de f.
[tex]f(x)=\sqrt{-x^2+2x+1}\\\\\\f'(x)=\dfrac{(-x^2+2x+1)'}{2\sqrt{-x^2+2x+1}}\\\\\\f'(x)=\dfrac{-2x+2}{2\sqrt{-x^2+2x+1}}\\\\\\f'(x)=\dfrac{2(-x+1)}{2\sqrt{-x^2+2x+1}}\\\\\\f'(x)=\dfrac{-x+1}{\sqrt{-x^2+2x+1}}\\\\\\\begin{array}{|c|ccccccccc|} x&-\infty&&1-\sqrt{2}&&1&&1+\sqrt{2}&&+\infty\\&&&&&&&&&\\-x+1&&+&+&+&0&-&-&-&\\\sqrt{-x^2+2x+1}&&||&0&+&0&+&0&||&\\&&&&&&&&&\\f'(x)&&||&||&+&0&-&||&||&\\&&&&&&&&&\\f(x)&&||&0&\nearrow&\sqrt{2}&\searrow&0&||&\end{array}[/tex]
soit
[tex]\begin{array}{|c|ccccc|} x&1-\sqrt{2}&&1&&1+\sqrt{2}\\&&&&&&&&&\\f(x)&0&\nearrow&\sqrt{2}&\searrow&0\end{array}[/tex]
