# Showing the existence of solutions

• Jan 18th 2010, 10:00 AM
alpha
Showing 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 AM
Jhevon
Quote:

Originally Posted by alpha
Show that the equation
x^2*e^mod(x)
Has at least two solutions for x in
-1<=x<=1.

Thanks

you do not have an equation here.
• Jan 18th 2010, 10:20 AM
alpha
Sorry. I made a typing error. Here is the corrected version.

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 AM
Plato
The expression $\displaystyle e^{\text{mod}(x)}$ is meaningless.
Should it be $\displaystyle e^{|x|}~?$
• Jan 18th 2010, 10:35 AM
alpha
Quote:

Originally Posted by Plato
The expression $\displaystyle e^{\text{mod}(x)}$ is meaningless.
Should it be $\displaystyle e^{|x|}~?$

Yes. That is what I mean. Sorry, I did not know how to get that symbol on here.

Thanks
• Jan 18th 2010, 10:37 AM
Jhevon
Quote:

Originally Posted by alpha
Sorry. I made a typing error. Here is the corrected version.

Show that the equation
x^2*e^mod(x)=2
Has at least two solutions for x in
-1<=x<=1.

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 AM
skeeter
Quote:

Originally Posted by alpha
Sorry. I made a typing error. Here is the corrected version.

Show that the equation
x^2*e^mod(x)=2
Has at least two solutions for x in
-1<=x<=1.

let $\displaystyle f(x) = x^2 \cdot e^{|x|} - 2$

$\displaystyle f(-1) = e-2 > 0$

$\displaystyle f(0) = -2 < 0$

$\displaystyle f(1) = e-2 > 0$

now use the Intermediate Value Theorem and the fact that f(x) is continuous over the given interval.
• Jan 18th 2010, 10:41 AM
Jhevon
Quote:

Originally Posted by alpha
Yes. That is what I mean. Sorry, I did not know how to get that symbol on here.

Thanks

it is on your keyboard. hold down shift and press the "\" key. | should show up

otherwise, type "abs" instead of "mod", "mod" means something else in math... but it's better and easier to use the | key
• Jan 18th 2010, 10:45 AM
alpha
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 AM
Jhevon
Quote:

Originally Posted by alpha
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.

what does the theorem say? try to apply it, try something! this "I have no idea how to..." line is starting to wear thin. try!
• Jan 18th 2010, 11:35 AM
alpha
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 AM
alpha
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 AM
skeeter
Quote:

Originally Posted by alpha
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 at least two solutions to f(x) = 0.

Is this correct?

...
• Jan 18th 2010, 11:57 AM
Jhevon
Quote:

Originally Posted by alpha
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 f(1) = f(-1) = e - 2 > 0, and f(0) = -2 < 0, f(x) changes sign between -1 and 0 and again between 0 and 1.

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

Is this correct?

yes! (Clapping)

you have the right idea. you just have to tighten it up a bit. you have to mention that the function is continuous and that you are applying the intermediate value theorem (twice), things like that.