Differentiability of $f(x)= 2-\mathrm{e}^{-x}$ when $x\geq 0$, $\mathrm{e}^{-x}$ otherwise

Question

I want to check for any $x_0$ in its domain, whether this function is differentiable or not. $f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \begin{cases} 2-\mathrm{e}^{-x} & ;\ x\geq 0 \\ \mathrm{e}^{-x} & ; x < 0 \end{cases}$

For $x \neq 0$ is $f$ obviously differentiable. Consider $x=0$

$\lim\limits_{x \downarrow 0}\frac{f(x)-f(0)}{x-0}= \lim\limits_{x \downarrow 0}\frac{2-\mathrm{e}^{-x}-2+\mathrm{e}^0}{x}=\lim\limits_{x \downarrow 0} \frac{-\mathrm{e}^{-x}+1}{x} \overset{L'Hôpital}{=} \lim\limits_{x \downarrow 0}\frac{\mathrm{e}^{-x}}{1}=1$

$\lim\limits_{x \uparrow 0}\frac{f(x)-f(0)}{x-0}= \lim\limits_{x \uparrow 0}\frac{\mathrm{e}^{-x}-1}{x} \overset{L'Hôpital}{=} \lim\limits_{x \uparrow 0}\frac{-\mathrm{e}^{-x}}{1}=-1$

Since right-hand and left-hand limit are not matching, $f$ is in $x_0=0$ not differentiable.

Could somebody please tell me, if this is correct?

just derive with the usual rules both expression, then take the limit! the first derivative is $e^{-x}$ which $ \to 1$; the second is $-e^{-x}$, which $\to -1$ — Ant, Jan 19 '14 at 23:33
On the right is continuos, and so is on the left. the only problem is in 0, so you can take the limit for $x \to 0^+$ and for $x \to 0^-$.. If this two limits are the same, then the derivative is continuos in 0. Otherwise (like in this case) it is not :-) — Ant, Jan 20 '14 at 00:17

score 1 · Answer 1 · answered Jan 20 '14 at 00:02

You are correct. Geometrically, you can see why. The graph is the graph of $x \mapsto e^{-x}$ with the part in the first quadrant reflected through the line $y=1$. This means the (nonhorizontal) tangent line at $x=0$ in the original graph becomes "broken" after the reflection since it would now abruptly change direction there. I.e., the left- and right-hand derivatives at $x=0$ do not agree, which you have shown analytically.

Differentiability of $f(x)= 2-\mathrm{e}^{-x}$ when $x\geq 0$, $\mathrm{e}^{-x}$ otherwise

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