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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 $\displaystyle |x|\cdot e ^ {\frac{1}{x-2}}=m$ 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 $\displaystyle m = |x|e^{-(x-2)}e^{(x-a)(x-b)(x-c)}$ $a,b,c$ 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, $\displaystyle \frac{x}{e^{(x-2)}} \quad x>0$ $\displaystyle \frac{-x}{e^{(x-2)}} \quad x<0$ $\displaystyle 0 \quad x=0$ 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: $\displaystyle \frac{d}{dx} \left( \frac{x}{e^{(x-2)}}\right) = \frac{e^{(x-2)}(1-x)}{e^{2(x-2)}}$ 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. $\displaystyle \lim_{x \to +\infty}\frac{x}{e^{(x-2)}} = 0$ 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 $x>0$ $\displaystyle \lim_{x \to -\infty}\frac{-x}{e^{(x-2)}} = \infty$ 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
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Joined: Aug 2018
From: România

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Quote:
 Originally Posted by Greens 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, $\displaystyle \frac{x}{e^{(x-2)}} \quad x>0$ $\displaystyle \frac{-x}{e^{(x-2)}} \quad x<0$ $\displaystyle 0 \quad x=0$ 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: $\displaystyle \frac{d}{dx} \left( \frac{x}{e^{(x-2)}}\right) = \frac{e^{(x-2)}(1-x)}{e^{2(x-2)}}$ 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. $\displaystyle \lim_{x \to +\infty}\frac{x}{e^{(x-2)}} = 0$ 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 $x>0$ $\displaystyle \lim_{x \to -\infty}\frac{-x}{e^{(x-2)}} = \infty$ 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, $m \in (0,e)$
Hello,

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
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From: USA

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Math Focus: Algebraic Geometry
Quote:
 Originally Posted by Integrator I do not understand your reasoning!
I misread the function as $|x| \cdot \frac{1}{e^{(x-2)}}$. The strategy I used was to bound the function and look at its behavior in different sections. Hopefully my misreading didn't mess you up, sorry.

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