Proving equation has only one solution
So i want to prove that $$x^2e^x=1$$ has at least one solution for $$x\in\mathbb{R}$$
I am kinda lost and would appreciate any help. This is suppose to be solved using basic calculus but i am not sure what to use.
$\endgroup$ 163 Answers
$\begingroup$Call $f(x)=x^2 e^x -1$, and note that it is continuous. Since $$\frac{d}{dx} (x^2 e^x -1) = e^x(x^2+2x)$$
we have that $(0,-1)$ is a maximum and $(-2,4e^{-2}-1)$ is a minimum.
Since $4e^{-2}-1<0$ and $\displaystyle \lim_{x\to -\infty} f(x)=-1$ we know that $f$ is negative in $(-\infty, -2)$. We also know that it is decreasing in $(-2,0)$, so still negative.
The function is strictly increasing in $(0, \infty)$ and since $\displaystyle \lim_{x\to\infty} f(x) = +\infty$ it has to cross the $x$ axis once (and only once) in that interval.
$\endgroup$ $\begingroup$Hint: Try to sketch a graph of $y=x^2 e^x$, paying particular attention to any minima and maxima it may have. Once you've done this, superimpose $y=1$ onto this graph to see what any solutions might look like.
1) Existence
If $f(x)=x^2e^x$, then
- $f$ is continuous
- $f(0)=0<1<e=f(1)$
By the intermediate value theorem, there exists a $c\in(0,1)$ such that $f(c)=1$, i.e. $c^2e^c=1$
2) Uniqueness
Using well-known inequality $e^y\ge1+y$ with $y=-1-\frac{x}{2}$, we see:
$$e^{-1-\frac{x}{2}}\ge -\frac{x}{2}\implies\frac{2}{e}\ge -xe^{x/2}\implies x^2e^x\le\frac{4}{e^2}<1\text{ for }x<0$$
and this tells us that $x^2e^x$ will have no solutions for negative $x$.
For positive $x$, $x^2e^x$ is monotonically increasing ($x^2,e^x$ are both positive and increasing, thus their product also is), so the positive solution $c$ we found earlier must be the only solution, and we thus have uniqueness.
$\endgroup$ $\begingroup$Since the statement is always positive $x^2$ is positive $e^x$ is positive we can take the logarithm and have more tangible:
$$x+2\log(|x|)=0$$
Now we have $\lim\limits_{x \to 0} x+2\log(|x|)=-\infty$. Observe that this is so for both sides of $0$: $0^{+}$ and $0^{-}$. So we are interested what is happening with the function once it leaves $0$.
Notice:
$$\lim\limits_{x \to +\infty} x+2\log(|x|)=+\infty$$
so there has to be at least one zero from $0$ to $+\infty$.
If we take first derivative of $f(x)=x+2\log(|x|)$ we have $f'(x)=\frac{x+2}{x}$
There is one zero of the first derivative at $x=-2$ for which $f(-2)=-2+2\log(2)<0$. This means that we cannot have zero in the region $x < 0$ since we would have to have at least one more minima/maxima somewhere within $x < 0$ and we do not.
For $x>0$ the expression $\frac{x+2}{x}$ is always positive, so the function is raising all the time from $-\infty$, and since it goes to $+\infty$ and the function has no other discontinuity except at $0$, we cannot have more than one value for which $f(x)=0$.
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