$f:D\to{R}$ $$f(x)=\frac{1}{x-2}e^{\left|x\right|}$$

Find the domain $D$ of the function and study whether the function is differentiable. Find the left and right derivatives in the points where the function isn't differentiable.

Since we have $x-2$ as the denominator I have to have $x\ne2$ which would give me $f\colon \mathbb{R}\setminus \{2\} \to \mathbb{R}$, that being the domain $D$ of the function.

As for differentiability, I write the function as follows:

$$f(x)=\left\{\begin{array}{cc} \frac{1}{x-2}e^x & x \ge 0 \\ \frac{1}{x-2}e^{-x} & x<0 \end{array}\right.$$

I then tried to find the left and right derivatives of the function in $x=0$ since I know that if they're equal, the function is differentiable.

I got $f'_s(0)=\frac{1}{4}$ and $f'_d(0)=-\frac{3}{4}$ which means the function isn't differentiable?

Was my thought process and work done right? I feel as if I have some kind of error. Also, shouldn't I check whether the function is differentiable in $x=2$ and find the left and right derivatives in that point as well?

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    $\begingroup$ Your work looks good to me. As you state, the function isn't even continuous at x = 2 (there isn't even some convenient value you can set f(2) to which would make it continuous), so differentiability at x = 2 isn't meaningful. $\endgroup$ – lulu Jul 7 '15 at 14:40
  • $\begingroup$ Yeah, but I have to calculate the side derivatives in points where the function isn't differentiable, so if it isn't at $x=2$ then I have to calculate them there too. Thanks for the reply! $\endgroup$ – MikhaelM Jul 7 '15 at 14:44
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    $\begingroup$ also your limites are ok, thus the function isn't differentiable at the point $x=0$ $\endgroup$ – Dr. Sonnhard Graubner Jul 7 '15 at 14:46
  • $\begingroup$ It seems that you only proved that the derivative is discontinuous at $0$. You have not shown that the derivative fails to exist. To do this, revert to the definition. All you really need to show is that the derivative of $|x|$ does not exits at $x=0$. If it did, then since composites of differentiable functions are differentiable, $e^{|x|}/(x-2)$ would be differentiable. If you don't like that, show that $e^{|x|}$ fails to be differentiable at $0$. Take $\lim_{h\to 0}\frac{e^{|h|}-1}{h}$ and see that you're back to taking $\lim_{h\to 0}\frac{|h|}{h}$, which does not exist. $\endgroup$ – Mark Viola Jul 7 '15 at 15:10
  • $\begingroup$ I didn't prove continuity or discontinuity, I think. I had limits of $f'(x)$ - the derivative, when $x\to0$ left and right, and I proved that those aren't equal, which according to what I know, would mean that the function isn't differentiable at x=0 ... From what I can tell the function is actually continuous at $x=0$ since $f_d(0) = f_s(0)=f(0)$ $\endgroup$ – MikhaelM Jul 7 '15 at 15:18

The requirement for a function to be differentiable is that it must be both smooth and continuous. Your function is discontinuous at x=2. So when you want to find the limit as x approaches 2 from the left and the right. Your function is continuous at x=0 so you're not going to look at the derivatives as x approaches zero from the left and right. You're close, the hint is in the question itself, it asks you to analyze where the function is no longer continuous, and then analyze the derivative (which is in essence taking the limit of the function over a delta x: by Definition) at the point where that discontinuity occurs.


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