# Math Help - Unique resolution

1. ## Unique resolution

Prove that equation , accepts an unique resolution : and :

2. Originally Posted by dhiab
Prove that equation , accepts an unique resolution : and :
put $g(x)=f(x)-x$, then:

$g'(x)=(-2x^2+x+3)e^{-x}-1$

Now look at $-2x^2+x+3$ this has roots at $x=-1$ and $3/2$, so $-2x^2+x+3$ has constant sign for $x>3/2$, a quick check shows that this is negative. Hence $g'(x)<0$ for $x>3/2$

Thus $g(x)$ is decreasing when $x>3/2$. To complete the proof it remains to show that $g(3/2)$ is positive and $g(2)$ is negative.

CB

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