Prove that the equation has an unique root Let $a>0$. The question is to show that the equation $ae^{x} - 1 = x +\frac{x^2}{2}$ has an unique real solution.
My attempt. Define $f(x) = ae^{x} - 1 - x - \frac{x^2}{2} $. We have $\lim_{x \rightarrow + \infty} f(x) = +\infty$ and $\lim_{x \rightarrow - \infty} f(x) = -\infty.$ Thus by the Intermediate value theorem there exist a root for the equation.
Regarding the uniqueness: My feeling is that it can be used the Rolle's Theorem but I am getting anywhere. someone could help me?
 A: Others have shown the uniqueness in the root for $a \ge 1$ by looking at the derivative. For $0<a<1$, the uniqueness can be found by looking at the relative maximum. The derivative of $f(x)$ is $$ae^{x}-1-x$$
Setting this equal to $0$ finds the relative maximum to be at $$x = -W(-\frac{a}{e})-1$$ where $W(x)$ is the Lambert W function.
The value of $f(x)$ at this $x$ is then $-\frac{(-W(-\frac{a}{e})-1)^2}{2}$. The derivative of the $y$ value of the relative maximum with respect to $a$ is then $$-\frac{W(-\frac{a}{e})}{a}$$
This is positive for $0<a<1$, and thus, the $y$ value of the relative maximum is increasing from $a = 0$ to $a = 1$. This relative maximum never equals $0$, only approaches it as $a$ approaches $1$. This claim is backed by the fact that, when $a = 1$, at the $x$-value where the derivative of $f(x)$ is $0$, the $y$-value is also $0$.
Now that it is shown that the relative maximum is never greater than or equal to $0$, it is trivial to show that there is only one distinct real root. Since the function increases until the relative maximum and still does not equal $0$, it will not equal $0$ at the relative minimum. Then, the Intermediate Value Theorem tells us there is a root and the derivative is positive. Therefore, there is only one real root for all values of $a$.
A: To see the uniqueness, check the sign of its derivative. If $f'(x)>0$ (which is clear) then $f$ is non-decreasing, which means that if you cut the $ x $-axis at some point, you cannot go down again.
