Let \(a>0\). Show that \[ ae^x=1+x+\frac{x^2}{2} \] has exactly one real solution.
Solution
Define \[ g(x)\coloneqq e^{-x}\left(1+x+\frac{x^2}{2}\right). \] The given equation is equivalent to \(g(x)=a\).
1. For every \(x\in\RR\),\[ g'(x)=-\frac{x^2}{2}e^{-x}. \]
Differentiating gives \[ \begin{aligned} g'(x) &=e^{-x}(1+x)-e^{-x}\left(1+x+\frac{x^2}{2}\right)\\ &=-\frac{x^2}{2}e^{-x}. \end{aligned} \]
2. The function \(g\) is strictly decreasing on \(\RR\).
If \(x<y\), the fundamental theorem of calculus and step 1 give \[ g(y)-g(x) = -\frac12\int_x^y t^2e^{-t}\,dt. \] The integrand \(t^2e^{-t}\) is nonnegative and is positive except at \(t=0\). Hence its integral over every nontrivial interval is strictly positive, so \(g(y)-g(x)<0\).
3. The endpoint limits of \(g\) are\[ \lim_{x\to-\infty}g(x)=+\infty, \qquad \lim_{x\to+\infty}g(x)=0. \]
Since \[ 1+x+\frac{x^2}{2} = \frac{(x+1)^2+1}{2} \geq \frac12, \] we have \(g(x)\geq \frac12e^{-x}\), which tends to \(+\infty\) as \(x\to-\infty\).
As \(x\to+\infty\), each of \(e^{-x}\), \(xe^{-x}\), and \(x^2e^{-x}\) tends to \(0\), so \(g(x)\to0\).
4. For every \(a>0\), there exists exactly one \(x\in\RR\) such that\[ g(x)=a. \]
By step 3, there exist real numbers \(u<v\) with \[ g(u)>a>g(v). \] The function \(g\) is continuous, so the intermediate value theorem gives some \(x\in(u,v)\) with \(g(x)=a\). Step 2 shows that \(g\) is strictly decreasing, so two distinct points cannot have the same value. Thus this solution is unique.
5. Q.E.D.
The original equation is equivalent to \(g(x)=a\), and step 4 proves that the latter equation has exactly one real solution.