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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?

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    $\begingroup$ For $a > 1$, you might notice that your $f(x)$ has a strictly positive derivative. For $a=1$, notice that $f'(x) = 0$ only at the origin and is positive everywhere else. For $0<a<1$, you could play around with the relative extrema. $\endgroup$ – Hyperion Jul 23 at 1:11
  • $\begingroup$ For $0<a<1$, there are two relative extrema. The relative max will always be decreasing as $a$ decreases from $a = 1$ to $a = 0$. Since the "relative max" at $a = 1$ is at $f(x) = 0$, the relative max will always be less than $0$ for $0<a<1$. $\endgroup$ – automaticallyGenerated Jul 23 at 1:27
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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 $a$.

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  • $\begingroup$ wow ! I think that my real analysis professor putted the wrong question in the exercises . thanks for your answer! $\endgroup$ – math student Jul 23 at 2:17
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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.

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  • $\begingroup$ We have $f^{'}(x) = ae^{x} - 1 -x .$ why $f^{'}(x) >0 $ for all x? $\endgroup$ – math student Jul 23 at 1:10
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    $\begingroup$ It depends of the choice of $a$, but a well-known inequality is the following : $$e^x \geq 1+x \quad \textrm{for all } x$$ So, the only thing that causes trouble is the case where $ 0 <a <1 $ $\endgroup$ – Azif00 Jul 23 at 1:12
  • $\begingroup$ This isn't true for $a \leq 1$ $\endgroup$ – Hyperion Jul 23 at 1:13
  • $\begingroup$ The inequality $e^{x} > 1 + x$ helps in the case $a \geq 1$ . Now the problem is the case $a <1$ $\endgroup$ – math student Jul 23 at 1:18

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