In a math problem I had to find the sign of the following :

$f(x) = -xe^{2x+1} - e^x$

I don't know how to prove it's sign, when it's negative or positive ?

I already know that $e^x > 0$ Same goes for $e^{2x+1}>0$ so it would follow the sign of $-x$ ( for $-xe^{2x+1}$ ) . However I can't find when is $f(x)$ positive or negative . Please help me out on this

Thank's for your time


You can simplify things by observing that your function can be written as $$ f(x)=e^x(-xe^{x+1}-1) $$ and studying the sign of $g(x)=-xe^{x+1}-1$ is just the same, because $e^x>0$.

The derivative is $$ g'(x)=-(e^{x+1}+xe^{x+1})=-e^{x+1}(1+x) $$ which only vanishes at $x=-1$. The derivative is negative for $x>-1$ and positive for $x<-1$, so $-1$ is a point of maximum (absolute, as there are no other critical points).

Since $g(-1)=e^0-1=0$, it follows that $g(x)<0$ for every $x\ne-1$.


My first thought about how to solve this is to factor.

$$-xe^{2x+1} - e^x = e^x(-xe^{x+1} -1)$$

Since $e^x$ is always positive we don't care about it. So

$$-xe^{x+1} -1 = 0$$

$$1= -xe^{x+1}$$

This equation is ugly and doesn't have any way of solving for $x$.

Luckily we can look use the wonderful method of guess and check to look for solutions. In particular $x=-1$ works and is the only time that $f(x)=0$.

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    $\begingroup$ $xe^x=c$ can be solved with the Lambert W function. It is $W(c)$ which in this case is $W(-1/e)=-1$. mathworld.wolfram.com/LambertW-Function.html $\endgroup$ – Hugh Nov 18 '16 at 21:27
  • $\begingroup$ @Hugh Cool. I haven't seen that before. However it's not easily solvable in terms that OP is familiar with. $\endgroup$ – lordoftheshadows Nov 19 '16 at 0:28

Using the functional equation of the exponential function $e^{x+y} = e^x e^y$, we can rewrite your function as $-e^x(xe^{x+1} + 1)$.

Now observe that $-e^x$ is negative for any value of $x$, so we only need to analyse the expression $xe^{x+1}+1$, which we will denote as $f(x)$ going forward.

$f(x)$ is obviously positive for any $x\ge0$, as $e^{x+1}$ is always positive and $x$ is positive in this case as well.

We consider the limit $\lim\limits_{x \rightarrow -\infty}{f(x)} = 1 = f(0)$. Thus, if there are to exist any $x$ where $f(x)$ is negative, then $f(x)$ must have a minimum in one of these points, as $f(x)$ is surely continuous.

Therefore, we now calculate $\frac{df(x)}{dx} = e^{x+1}(x+1)$. This is only zero for $x_0 = -1$, which the second derivative test indeed reveals to be a minimum of $f(x)$.

But $f(x_0) = 0$, therefore there are no values of $x$ for which $f(x)$ is negative.

As your initial function was defined as $-e^x f(x)$, we can conclude that your function is negative for any real $x$ except for $x_0=-1$, where your function takes the value $0$.

  • $\begingroup$ So this is basically what egreg wrote. He posted his answer while I was writing mine - do I delete this now? I'm not sure what the appropriate way to handle this is. $\endgroup$ – Tom Nov 18 '16 at 21:37

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