Réponse : Bonjour,
Partie B
3) [tex]g(\alpha)=0[/tex], donc :
[tex]2e^{\alpha}+2\alpha-5=0\\2e^{\alpha}=-2\alpha+5\\e^{\alpha}=\frac{-2\alpha+5}{2} \\\\\frac{1}{e^{\alpha}} =\frac{2}{-2\alpha+5} \\e^{-\alpha}=\frac{2}{-2\alpha+5}[/tex].
Puis:
[tex]f(\alpha)=(2\alpha-3)(1-e^{-\alpha})\\f(\alpha)=(2\alpha-3)(1-\frac{2}{-2\alpha+5} )\\f(\alpha)=(2\alpha-3)(\frac{-2\alpha+5-2}{-2\alpha+5} )\\f(\alpha)=(2\alpha-3)(\frac{-2\alpha+3}{-2\alpha+5} )\\f(\alpha)=(2\alpha-3)(\frac{-(2\alpha-3)}{-(2\alpha-5)} )\\f(\alpha)=(2\alpha-3)(\frac{2\alpha-3}{2\alpha-5} )\\f(\alpha)=\frac{(2\alpha-3)^{2}}{2\alpha-5}[/tex].
4) La fonction [tex]h(x)= \frac{(2x-3)^{2}}{2x-5}[/tex] est croissante, donc:
[tex]h(0,627)< h(\alpha) < h(0,628)\\h(0,627)< f(\alpha) < h(0,628) \quad car \: f(\alpha)=h(\alpha)\\-0,814 < f(\alpha) < -0,812[/tex].
