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April 21st, 2019, 09:55 PM  #1 
Senior Member Joined: Aug 2018 From: România Posts: 108 Thanks: 7  An equation with a parameter
Hello all, For which values of the parameter $\displaystyle m$ does the equation $\displaystyle x\cdot e ^ {\frac{1}{x2}}=m$ have three real and distinct solutions? All the best, Integrator Last edited by skipjack; April 22nd, 2019 at 10:59 AM. 
April 22nd, 2019, 06:26 AM  #2 
Senior Member Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry 
$\displaystyle m = xe^{(x2)}e^{(xa)(xb)(xc)}$ $a,b,c$ unequal. 
April 22nd, 2019, 09:38 AM  #3 
Senior Member Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry 
I realize you're most likely looking for scalar $m$, apologies. You can take the derivative and solve for $0$ to find the local max. Since the function isn't differentiable at $x=0$, you can split the function into three parts, $\displaystyle \frac{x}{e^{(x2)}} \quad x>0$ $\displaystyle \frac{x}{e^{(x2)}} \quad x<0$ $\displaystyle 0 \quad x=0$ We need to find what $y$ values the function captures three times, so it's helpful to find some bounds. We can take the derivative and solve for $\frac{d}{dx} = 0$ to find critical values: $\displaystyle \frac{d}{dx} \left( \frac{x}{e^{(x2)}}\right) = \frac{e^{(x2)}(1x)}{e^{2(x2)}}$ So $x=1$ is a critical value, $y=e$ at that point. This is a maximum. For a lower bound, we take the limit as $x \to \infty$. Since the function is continuous for $x>0$, we can get some information from it. $\displaystyle \lim_{x \to +\infty}\frac{x}{e^{(x2)}} = 0$ So, now we know the function intersects the values in $y=(0,e)$ twice, once in $x\in (0,1)$ and again in $x\in(1,+\infty)$ now we check $x<0$. We can check the limit for $x<0$ to see if the functions intersects $y \in (0,e)$ using the same method as $x>0$ $\displaystyle \lim_{x \to \infty}\frac{x}{e^{(x2)}} = \infty$ So, since the function has a value for every $x<0$, and the limit goes to $+\infty$, the function must intersect $y \in (0,e)$ once in $x<0$. Since the function only has one critical value (besides $x=0$) there are only 3 distinct intersections, once in $x<0$, once in $x\in (0,1]$ and again in $x\in(1,+\infty)$. So, $m \in (0,e)$ Last edited by skipjack; April 22nd, 2019 at 10:59 AM. Reason: latex errors 
April 28th, 2019, 10:27 PM  #4  
Senior Member Joined: Aug 2018 From: România Posts: 108 Thanks: 7  Quote:
I do not understand your reasoning!The correct answer is $\displaystyle m\in (0,\frac{1}{e})\cup (4\sqrt{e} , +\infty)$ All the best, Integrator  
April 29th, 2019, 08:32 AM  #5 
Senior Member Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry  

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