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April 21st, 2019, 09:55 PM   #1
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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.
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April 22nd, 2019, 06:26 AM   #2
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$\displaystyle m = |x|e^{-(x-2)}e^{(x-a)(x-b)(x-c)}$

$a,b,c$ unequal.
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April 22nd, 2019, 09:38 AM   #3
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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
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April 28th, 2019, 10:27 PM   #4
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Quote:
Originally Posted by Greens View Post
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
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April 29th, 2019, 08:32 AM   #5
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Quote:
Originally Posted by Integrator View Post

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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