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

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January 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 :)
January 13th 2010, 06:02 AM
Dinkydoe

If you may use that $f(x) = x^2e^{|x|}-2$ is continuous, it shouldn't be that hard: $f[a,b]\to \mathbb{R}$ must reach all values in $[f(a), f(b)]$ .

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

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

You can use this argument twice to show the existence of 2 roots (without necessarily finding them).
January 13th 2010, 06:12 AM
Dinkydoe

I meant the Intermediate value Theorem. It says: If f is contunious on a closed interval $f:[a,b]\to \mathbb{R}$ then for any $c\in [f(a),f(b)]$ exists a $x_0\in [a,b]$ such that $f(x_0)=c$ .

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