Show that the equation

x^2*e^mod(x)

Has at least two solutions for x in

-1<=x<=1.

Thanks

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- Jan 18th 2010, 10:00 AMalphaShowing the existence of solutions
Show that the equation

x^2*e^mod(x)

Has at least two solutions for x in

-1<=x<=1.

Thanks - Jan 18th 2010, 10:02 AMJhevon
- Jan 18th 2010, 10:20 AMalpha
Sorry. I made a typing error. Here is the corrected version.

Thank you for your help.

Show that the equation

x^2*e^mod(x)=2

Has at least two solutions for x in

-1<=x<=1. - Jan 18th 2010, 10:33 AMPlato
The expression $\displaystyle e^{\text{mod}(x)}$ is meaningless.

Should it be $\displaystyle e^{|x|}~?$ - Jan 18th 2010, 10:35 AMalpha
- Jan 18th 2010, 10:37 AMJhevon
Note that this is the same as showing $\displaystyle x^2 e^{|x|} - 2 = 0$ at least twice in the given range.

Let $\displaystyle f(x) = x^2 e^{|x|} - 2$ (note that it is a continuous function) and consider $\displaystyle f(-1),~f(0), \text{ and }f(1)$ and apply the intermediate value theorem. - Jan 18th 2010, 10:39 AMskeeter
- Jan 18th 2010, 10:41 AMJhevon
- Jan 18th 2010, 10:45 AMalpha
I have no idea of how to apply this to the intermediate value theorem. Can you show me how to answer this question.

Thank you. - Jan 18th 2010, 10:47 AMJhevon
- Jan 18th 2010, 11:35 AMalpha
Well I ahve found out that the theorem is as follows:

If ƒ(*x*) is a continuous real-valued function on the closed interval from*a*to*b*, then, for any*y*between the least upper bound and the greatest lower bound of the values of ƒ, there is an*x*between*a*and*b*with ƒ(*x*) =*y*

Now how do I apply this to the question. - Jan 18th 2010, 11:45 AMalpha
I think i have just worked it out.

For f(x) = x^2 * e^abs(x) -2 f(-1) = e - 2 and f(0) = -2 and f(1) = e - 2

Since e - 2 > 0, f(x) changes sign between -1 and 0 and again between 0 and 1.

So there are two solutions to f(x) = 0.

Is this correct? - Jan 18th 2010, 11:56 AMskeeter
- Jan 18th 2010, 11:57 AMJhevon