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July 1st, 2014, 08:50 PM   #1
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Finding the roots (or critical numbers)

Hello,

I have taken to derivative of an equation to be e^(-x^2) (1-2 x^2). I am trying to find the critical numbers so I can do test points etc. and figure out where the graph of f is decreasing/increasing etc.

I thought I knew how to figure it out but keep getting hung up. I believe the roots should be +/- (1/√2)

Thanks for your help.
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July 2nd, 2014, 02:11 AM   #2
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I am not sure what you're asking for but I believe you're asking for the extremum and then determine whether the function is going up or down, right?

You have an extremum whenever your derivative is 0 (and your second derivative isn't 0 but that's not very important)
and that is indeed at +/- (1/√2)
Before the first extremum the function is going down, then inbetween the function is going upwards and after the second extremum it's going down again.

Is this satisfying or do you need to know something else?

Last edited by skipjack; November 4th, 2016 at 12:33 AM.
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July 2nd, 2014, 02:45 AM   #3
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Sorry, I could have done a better job of explaining. I am confused with the algebra that leads to getting +/- (1/√2). I am getting lost in the algebra after I set e^(-x^2) (1-2 x^2) = 0.
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July 2nd, 2014, 03:01 AM   #4
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Quote:
Originally Posted by OrangeT View Post
I am getting lost in the algebra after I set e^(-x^2) (1-2 x^2) = 0.
So we are trying to solve $\displaystyle e^{-x^2} * (1-2 x^2) = 0$
This means that either $\displaystyle e^{-x^2} = 0$ or $\displaystyle (1-2 x^2) = 0$.

If we try to solve $\displaystyle e^{-x^2} = 0$ we should take the natural logarithm of both sides but since $\displaystyle \ln(0)$ doesn't exist this equation has no solutions.

Then we try to solve $\displaystyle (1-2 x^2) = 0$. We bring the $\displaystyle 2x^2$ term to the other side of the equation to get $\displaystyle 1=2 x^2$. We now divide both sides by 2: $\displaystyle \frac {1}{2} = x^2$. Now we just have to take the square root of both sides and we will have our desired result. $\displaystyle \pm \sqrt {\frac {1}{2} } = x$ or simply $\displaystyle \pm \frac {1}{\sqrt{2}} = x$
The answer can be negative or positive because as you can see if you square it again the sign will fall away anyway.

Last edited by skipjack; November 4th, 2016 at 12:34 AM.
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July 2nd, 2014, 05:46 PM   #5
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Agh, thank you. I guess I wasn't realizing I could set both sides in e−x2∗(1−2x2) equal to zero. Thank you for the help!
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