Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
    
Thank you very much! –  Bilbo Dec 16 '12 at 14:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.