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It is required to analytically(without any online tool) find the number solutions to the equation $3xe^x+1=0$.

As there is no solution at 0, the problem is $$e^x=-\frac {1}{3x}$$ at $x=-1$,$e^x$ is below the other($e<3$), but also means that the slope of decrease of $e^x$ is still sharper.So it not determinable yet.

What I did was to modify the problem by substituting $x$ for $-t$ and then flipping the equation (which is possible because there is an equality) which yields $e^x=3x$ , which quite clearly has 2 solutions.

My question is how should we analyse the above two functions 'purely' (that is by analysing there own slopes and behaviour) for a more general method to find the number of solutions. My concern is for the more general problems where such simple tricks won't work and rigorous analysis would be needed.

EDIT: to find the number of solutions $e^x=3x$, Let us start by assuming a $mx$ which should be tangential to $e^x$ at some $x=a$,then $$e^a=ma$$ and by their slopes $$e^a=m$$ dividing the two yields $$a=1$$ and hence $m=e$ .Then as $3>e$ ,there will be two roots for $3x$.

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    $\begingroup$ $f'(x)=3(x+1)e^x$ tells you that the function is strictly decreasing for $x<-1$ and strictly increasing for $x>-1$. Now consider the limits of $f(x)$ for $x\to -\infty$ and $x\to \infty$ and $f(-1)<0$ $\endgroup$
    – Peter
    Apr 27 at 11:44
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    $\begingroup$ Your approach is creative (however why has $e^x=3x$ "clearly" two solutions ? What is your agument for this ?) , but in general consider the monotony intervals. $\endgroup$
    – Peter
    Apr 27 at 11:51
  • $\begingroup$ @Peter ,Thank you very much ! The analysis of $3xe^x$ helps(the graph 'dips' below y=-1 and tends to 0- at $-\inf$ ) Also edited as suggested. $\endgroup$ Apr 27 at 12:28

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As noted in the comments, "a miracle occurs" when the equation $e^t=3t$ "clearly has two solutions".

One way to fill this gap involves the use of Descartes' Rule of Signs. Ordinarily this is applied to polynomials, but it extends to infinite Taylor series if the series converges and has only finitely many sign changes.

To wit, render

$e^t-3t=0$

$1\color{blue}{-}2t\color{blue}{+}\frac12 t^2+\frac16 t^3+ ...=0$

The higher degree terms, coming from the series for $e^t$, all have positive coefficients and so there are just the two sign reversals indicated by the blue signs above. Hence at most two positive roots for $t$.

The existence of this pair of solutions is then proven by noting that $e^t-3t<0$ at $t=1$. These roots then correspond to negative solutions for $x=-t$. Add the fact that there are clearly no nonnegative real solution for $x$ in the original equation and we are done.

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    $\begingroup$ +1 this is a great way to see that . Noted. $\endgroup$ Apr 27 at 12:39
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Use the well-known special function Lambert W :

For real number $x,y$ the equation

$ye^y=x$ can be solved for $y$, if and only if $x≥-\frac 1e$.

We have

$y=W_0(x)$ if $x≥0$ and the two values

$y=W_0(x)$ and $y=W_{-1}(x)$ if $-\frac 1e≤x<0$.

Since $e<3$, we get

$$-\frac 1e ≤-\frac 13<0$$

then the answer is obvious.

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  • $\begingroup$ Thank you for your time. A little off topic question from me, Is this special function on an undergrad level course? ( I am in high school so hadn't heard of it before) $\endgroup$ Apr 27 at 13:13
  • $\begingroup$ In high school we learn only elementary functions. @Sungjin-woo $\endgroup$
    – User
    Apr 27 at 13:54
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As $ \ e^x \ $ is positive everywhere, any root of $ \ f(x) \ = \ e^x \ = \ g(x) \ = \ -\frac{1}{3x} \ $ must be $ \ x \ < \ 0 \ \ . $ As both functions are continuous over $ \ x \ < \ 0 \ \ , $ we may apply the Intermediate Value Theorem. We observe that

$ • \ \ \ e \ < \ 3 \ \ \Rightarrow \ \ g(-1) \ = \ -\frac{1}{3·(-1)} \ = \ \frac13 \ < \ \frac{1}{e} \ = \ e^{-1} \ = \ f(-1) \ \ ; $

$ • \ \ \ e \ > \ \frac94 \ \ \Rightarrow \ \ \frac{1}{e} \ < \ \frac49 \ \ \Rightarrow \ \ f \left(-\frac12 \right) \ = \ e^{-1/2} \ = \ \frac{1}{\sqrt{e}} \ < \ \sqrt{\frac49} \ = \ \frac23 \ = \ -\frac{1}{3·(-1/2)} \ = \ g \left(-\frac12 \right) \ \ ; $

$ • \ \ \ 2.6^2 \ = \ 6.76 \ > \ 6 \ \ \Rightarrow \ \ e \ > \ 2.6 \ > \ \sqrt6 \ \ \Rightarrow \ \ f(-2 ) \ = \ e^{-2} \ < \ \frac{1}{6} \ = \ -\frac{1}{3·(-2)} \ = \ g(-2 ) \ \ $

[this last result is not straightforward to establish using simple inequalities; I resorted to a numerical "reference" here].

Hence there are roots $ \ -2 \ < \ x_1 \ < \ -1 \ $ and $ \ -1 \ < \ x_2 \ < \ -\frac12 \ \ $ for the equation $ \ e^x \ = \ -\frac{1}{3x} \ \ . $

[It is also not easy to "narrow" the interval for $ \ x_1 \ $ without computational aid, since $ \ x_1 \ $ is very close to $ \ -\frac32 \ . \ $ ]

enter image description here

Are there just these two roots? With $ \ f'(x) \ = \ e^x \ \ , \ \ g'(x) \ = \ \frac{1}{3x^2} \ \ , $ we can show that

$ • \ \ \ e^{x_1} \ = \ -\frac{1}{3x_1} \ \ \Rightarrow \ \ f'(x_1) \ = \ e^{x_1} \ > \ \frac{1}{3x_1^2} \ = \ g'(x_1) \ \ , \ $ since $ \ |x_1| \ > \ 1 \ \ ; $

$ • \ \ \ g'(-1) \ = \ \frac{1}{3·(-1)^2} \ = \frac13 \ < \ \frac{1}{e} \ = \ e^{-1} \ = \ f'(-1) \ \ ; $

$ • \ \ \ e^{x_2} \ = \ -\frac{1}{3x_2} \ \ \Rightarrow \ \ f'(x_2) \ = \ e^{x_2} \ < \ \frac{1}{3x_2^2} \ = \ g'(x_2) \ \ , \ $ since $ \ |x_2| \ < \ 1 \ \ . $

So $ \ -\frac{1}{3x} \ $ increases more slowly than $ \ e^x \ $ over most of the interval $ \ (x_1 \ , \ x_2) \ \ , $ only "growing faster" the exponential function near $ \ x_2 \ $ and thereafter as $ \ x \ = \ 0 \ $ is "approached from below." (We find that $ \ f'(x) \ = \ g'(x) \ $ at $ \ x \ \approx \ -0.91 \ . \ ) $ Hence, returning to the original equation, $ \ 3xe^x + 1 \ = \ 0 \ $ has only two real roots.

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Consider that you look for the zeros of function $$f(x)=3x\,e^x+1$$ the first derivative cancels at $x_*=-1$; we have $f(-1)=1-\frac{3}{e}<0$ and $f''(x_*)=\frac{3}{e}>0$. So, there is at least two roots (on each side of $-1$).

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