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Could somebody show me how to prove that $ ae^x=1+x+\frac{x^2}{2}$ has only one real root for $a>0$? All I know so far is that the equation has a root because $1+x+\frac{x^2}{2}>0$ for all real x.

There can't be two solutions, because the quadratic function is always above x axis, and so is $e^x$. When I drew a graph of those two functions I found that the only point of intersection is 0. I wish I could include the graph here but since I'm new here, my reputation is too low and I can't.

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The equation is true for all $x$ and you want to find $a$ or the opposite? – Nameless Dec 16 '12 at 14:36
@Amr: Three, I think. – Henning Makholm Dec 16 '12 at 14:47
@ Henning Makholm No this one does not work as well. – Amr Dec 16 '12 at 14:54
up vote 1 down vote accepted

Define $f(x) = ae^x - 1 - x -x^2/2.$ As $x\to -\infty$, $f(x) \to -\infty$ and as $x\to \infty$, $f(x) \to \infty,$ and also $f$ is continuous, so by the Intermediate Value theorem, $f$ has at least $1$ root.

Now suppose $x_0$ is a point such that $f(x_0)$ is a maximum or a minimum. We have $f'(x) = ae^x-1-x$ and since $f'(x_0)=0,$ we get $f(x_0) = -x_0^2/2.$ Thus every extremum occurs when $f$ is negative, so $f$ has precisely $1$ root.

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Thank you very much! – Bilbo Dec 16 '12 at 14:57

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