Find the extrema of f(x) = cases e^(-1/|x|)(sqrt2+ sin

Find the extrema of $f(x) = \begin{cases} e^{-1/|x|}\left(\sqrt2+\sin\dfrac1x\right), & x\neq0,\\[6pt] 0, & x = 0. \end{cases}$.

Step 1: The function is non-negative

For $x\neq0$ the exponential is positive and $\sqrt2+\sin\dfrac1x\ge\sqrt2-1\gt 0$, so $f(x)\gt 0$; and $f(0) = 0$. Thus

\[ f(x)\ge0 \text{ everywhere, with equality only at } x = 0. \tag{1} \]

That already settles the answer — but the derivative shows how the function approaches it.

Step 2: Continuity and differentiability at the origin

Since $e^{-1/|x|}\to0$ while the bracket stays between $\sqrt2-1$ and $\sqrt2+1$, $\lim_{x\to0}f(x) = 0 = f(0)$, so $f$ is continuous there. Moreover \[ f'(0) = \lim_{x\to0}\frac{f(x)-0}{x} = \lim_{x\to0}\frac{e^{-1/|x|}}{x}\left(\sqrt2+\sin\frac1x\right) = 0, \] because $\dfrac{e^{-1/|x|}}{|x|}\to0$ — exponential decay beats any power — and the bracket is bounded.

Step 3: The derivative for $x\gt 0$

\[ f'(x) = \frac{e^{-1/x}}{x^{2}}\left(\sqrt2+\sin\frac1x\right)-\frac{e^{-1/x}}{x^{2}}\cos\frac1x = \frac{e^{-1/x}}{x^{2}}\left(\sqrt2+\sin\frac1x-\cos\frac1x\right). \tag{2} \]

With $t = \dfrac1x$ and $\sin t-\cos t = \sqrt2\sin\bigl(t-\tfrac{\pi}{4}\bigr)$, \[\sqrt2+\sin t-\cos t = \sqrt2\left(1+\sin\left(t-\frac{\pi}{4}\right)\right)\ge0, \tag{3}\] so $f'\ge0$: the function is non-decreasing on $\bigl(0,\infty\bigr)$. Equality holds only where $\sin\bigl(\tfrac1x-\tfrac{\pi}{4}\bigr) = -1$, that is at the isolated points $x_{k} = \dfrac{4}{\bigl(8k-1\bigr)\pi}$, $k\ge1$ — stationary points at which $f'$ does not change sign, hence not extrema.

Step 4: The derivative for $x\lt 0$

There $|x| = -x$, so $f(x) = e^{1/x}\bigl(\sqrt2+\sin\tfrac1x\bigr)$ and

\[ f'(x) = -\frac{e^{1/x}}{x^{2}}\left(\sqrt2+\sin\frac1x+\cos\frac1x\right) = -\frac{e^{1/x}}{x^{2}}\cdot\sqrt2\left(1+\sin\left(\frac1x+\frac{\pi}{4}\right)\right)\le0. \tag{4} \]

So $f$ is non-increasing on $\bigl(-\infty,0\bigr)$.

Step 5: The conclusion about extrema

The function falls (weakly) all the way down to $x = 0$ from the left and rises (weakly) from $x = 0$ to the right, and it is strictly positive on both sides. Hence $x = 0$ is a strict global minimum with $f(0) = 0$.

As $|x|\to\infty$ we have $e^{-1/|x|}\to1$ and $\sin\dfrac1x\to0$, so $f(x)\to\sqrt2$, and by monotonicity $f(x)\lt \sqrt2$ throughout. The value $\sqrt2$ is a supremum that is never attained: there is no maximum.

Step 6: Verification

\[f(0.05) = 4.8\cdot10^{-9} \lt f(0.2) = 0.0030677 \lt f(1) = 0.8298200 \lt f(100) = 1.4100422 \;\;\checkmark\]

\[f(-1) = 0.2107002 \gt \ldots;\quad f(-5) = 0.9952034,\ f(-100) = 1.3902416 \;\;\checkmark\ (\text{rising as } x\to-\infty)\]

\[\sqrt2 = 1.4142136 \text{ is never reached} \;\;\checkmark\]