
Unique resolution

Quote:
Originally Posted by
dhiab
put $\displaystyle g(x)=f(x)x$, then:
$\displaystyle g'(x)=(2x^2+x+3)e^{x}1$
Now look at $\displaystyle 2x^2+x+3$ this has roots at $\displaystyle x=1$ and $\displaystyle 3/2$, so $\displaystyle 2x^2+x+3$ has constant sign for $\displaystyle x>3/2$, a quick check shows that this is negative. Hence $\displaystyle g'(x)<0$ for $\displaystyle x>3/2$
Thus $\displaystyle g(x)$ is decreasing when $\displaystyle x>3/2$. To complete the proof it remains to show that $\displaystyle g(3/2)$ is positive and $\displaystyle g(2)$ is negative.
CB

Hello captain thank you for your resolution (Clapping)Ajout : alpha = 1.91 and g(1.5)=0.5 , g(2)=0.12 (Angry)