Real Analysis Real Analysis Math Forum

 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
Senior Member

Joined: Aug 2018
From: România

Posts: 108
Thanks: 7

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
Senior Member

Joined: Oct 2018
From: USA

Posts: 102
Thanks: 77

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. Tags equation, parameter Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Alvalance789 Algebra 8 November 27th, 2018 03:51 AM goldgold Algebra 5 December 26th, 2014 06:15 AM Dacu Algebra 4 August 10th, 2013 10:37 PM 940108 Calculus 5 May 21st, 2013 04:50 AM mathkid Algebra 6 August 5th, 2012 09:56 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      