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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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1 Answer 1

up vote 2 down vote accepted

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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