# Show that x^2*e^|x|=2 has two solutions

• Jan 13th 2010, 04:04 AM
DBsHo
Show that x^2*e^|x|=2 has two solutions
How would I do this? I know from the graph that the roots are -0.9 and 0.9 but how do I show this??? Any help much appreciated.
Thank you :)
• Jan 13th 2010, 06:02 AM
Dinkydoe
If you may use that $\displaystyle f(x) = x^2e^{|x|}-2$ is continuous, it shouldn't be that hard: $\displaystyle f[a,b]\to \mathbb{R}$ must reach all values in $\displaystyle [f(a), f(b)]$.

(look up theorem of Bolzano. I believe there's other even other names for that)

given that $\displaystyle f(1) = e-2 > 0$ and $\displaystyle f(1/2) < 0$ it follows that there exists a $\displaystyle x_0$ with $\displaystyle 1/2 < x_0 < 1$ such that $\displaystyle f(x_0) = 0$.

You can use this argument twice to show the existence of 2 roots (without necessarily finding them).
• Jan 13th 2010, 06:12 AM
Dinkydoe
I meant the Intermediate value Theorem. It says: If f is contunious on a closed interval $\displaystyle f:[a,b]\to \mathbb{R}$ then for any $\displaystyle c\in [f(a),f(b)]$ exists a $\displaystyle x_0\in [a,b]$ such that $\displaystyle f(x_0)=c$.