My Math Forum Finding the roots (or critical numbers)

 Algebra Pre-Algebra and Basic Algebra Math Forum

 July 1st, 2014, 08:50 PM #1 Member   Joined: Apr 2014 From: Japan Posts: 37 Thanks: 3 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.
 July 2nd, 2014, 02:11 AM #2 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 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.
 July 2nd, 2014, 02:45 AM #3 Member   Joined: Apr 2014 From: Japan Posts: 37 Thanks: 3 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.
July 2nd, 2014, 03:01 AM   #4
Senior Member

Joined: Mar 2012
From: Belgium

Posts: 654
Thanks: 11

Quote:
 Originally Posted by OrangeT 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.

 July 2nd, 2014, 05:46 PM #5 Member   Joined: Apr 2014 From: Japan Posts: 37 Thanks: 3 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!

 Tags critical, finding, numbers, roots

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post SamFe Calculus 2 November 6th, 2013 11:11 AM steelcurtain11 Calculus 2 November 1st, 2011 06:51 PM TsAmE Complex Analysis 4 October 19th, 2010 08:48 PM HFH Algebra 3 May 29th, 2010 01:19 PM nisko Complex Analysis 2 November 6th, 2008 08:52 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top