# For a fixed and small $\epsilon$, finding the number of real roots of $x^{2}+e^{-\epsilon x}-2+\sin(\epsilon x)$

I saw the following question in an introduction to applied mathematics exam (this is only the first part of the question):

Assume $0<\epsilon\ll1$ . Denote $$f(x,\epsilon):=x^{2}+e^{-\epsilon x}-2+\sin(\epsilon x)$$

How many real roots does $f(x,\epsilon)$ have ?

The answer stated that:

$$f(0,\epsilon)=1-2=-1$$ $$f_{x}(0,\epsilon)=0$$

hence $0$ is an extremum. Since $$f_{xx}(0,\epsilon)>2$$

we have it that $0$ is a minimum.

Now, since this function is increasing for a positive $x$, for a small enough $\epsilon$, and $f(x,\epsilon)\to\infty$ as $x\to\infty$ than there is exactly one positive root.

A similar argument show that there is also exactly one negative root.

My main problem with this answer is that I don't understand why for a positive $x$ the function is increasing : its easy to see that $x^{2}-2$ is increasing, but the other two elements $e^{-\epsilon x},\sin(\epsilon x)$ are not.

Can someone please explain the claim that $f(x,\epsilon)$ is increasing for a positive $x$, and/or suggest an alternative method for solving this question ?

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## 2 Answers

Hints:

For very small $\;\epsilon >0\;$ we have that

$$x^2+e^{-\epsilon x}-2+\sin\epsilon x \cong x^2+1-2+0=x^2-1$$

and the last expression on the right indeed has two zeros, one positive and one negative...and no need to go into two variable functions, partial derivatives and stuff, but:

$$f'(x)=2x-\epsilon e^{-\epsilon x}+\epsilon\cos\epsilon x\cong 2x-\epsilon+\epsilon = 2x>0\;,\;\;\text{for very small}\;\epsilon>0\ldots$$

and this is the reason they say the function's increasing for $\,x>0\,$

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DonAntonio's answer isn't quite correct. When your first approximation in terms of $\epsilon$ cancels out, it doesn't mean that the values are equal: it means that you need to look at the next higher power of $\epsilon$.

So: $$-\epsilon e^{-\epsilon x} = -\epsilon + \epsilon\frac{\epsilon x}{1!} -\epsilon\frac{\epsilon^2 x^2}{2!} +…$$ and: $$\epsilon\cos {\epsilon x} = \epsilon - \epsilon\frac{\epsilon^2 x^2}{2!}+...$$

…and their sum is not 0 (as DonAntonio stated) but $\epsilon^2-\epsilon^3 x^2$, which, for $\epsilon$ small enough, is dominated by $\epsilon^2$. This is positive, which means that $f'(x)>2x$, which is what is needed for the result.

The result is therefore correct, but the argument was fallacious because it missed this extra point - and in other circumstances this could lead to an erroneous result.

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