Which points in the domain of f are differentiable and what is their derivative?

[mk9898]

Homework Statement

Let f be $f:[0,\infty]\rightarrow \mathbb R \\ f(x):= \begin{cases} e^{-x}sin(x), \ if \ \ x\in[2k\pi,(2k+1)\pi] for \ a \ k \in \mathbb N_0 \\ 0 \ \ otherwise\\ \end{cases}$

Exercise: Determine all inner points of the domain where f is also differentiable and determine f' at these points.

The Attempt at a Solution

Idea:

Since we only need to look at inner differentiable points, we can look at points all points where f'(x) = 0. These are the only differentiable points because this is where the second case of f is a constant i.e. f'(x') = 0 for all x' in the second case. So we need to look for the points in the first case, where f'(x'')=0 and x'' is in these closed intervals for the first case. These are the differentiable inner points.

If I am on the right track, then I am wondering how I could find all of the points where that is valid for the first case of f.

 

[tnich]

Homework Statement

Let f be $f:[0,\infty]\rightarrow \mathbb R \\ f(x):= \begin{cases} e^{-x}sin(x), \ if \ \ x\in[2k\pi,(2k+1)\pi] for \ a \ k \in \mathbb N_0 \\ 0 \ \ otherwise\\ \end{cases}$

Exercise: Determine all inner points of the domain where f is also differentiable and determine f' at these points.

The Attempt at a Solution

Idea:

Since we only need to look at inner differentiable points, we can look at points all points where f'(x) = 0. These are the only differentiable points because this is where the second case of f is a constant i.e. f'(x') = 0 for all x' in the second case. So we need to look for the points in the first case, where f'(x'')=0 and x'' is in these closed intervals for the first case. These are the differentiable inner points.

If I am on the right track, then I am wondering how I could find all of the points where that is valid for the first case of f.

By "inner point" do you mean an interior point of the domain, i.e., not on a boundary?
I am not sure what you mean by $N_0$ either. I am assuming whole numbers.

Anyway, it appears that the function is defined for non-negative reals so a point x is interior if x > 0. All you need to do is figure out at what points $lim_{x\to0^+}f'(x)\neq lim_{x\to0^-}f'(x)$. All other x > 0 are differentiable points.

 

[mk9898]

Yes inner point = interior point. N_0 is natural numbers + 0.

I thought that all points that fulfill $\lim_{x \to 0} f'(x) = 0$ are the differentiable points?

 

[mk9898]

Could you also explain, why I should only consider as x ->0 and not other values?

 

[tnich]

Could you also explain, why I should only consider as x ->0 and not other values?

Sorry, I wrote that incorrectly. What I meant to say was, the non-differentiable points will be the values $x'$ such that
$lim_{x\to x'^+}f'(x)\neq lim_{x\to x'^-}f'(x)$.

 

[tnich]

Sorry, I wrote that incorrectly. What I meant to say was, the non-differentiable points will be the values $x'$ such that
$lim_{x\to x'^+}f'(x)\neq lim_{x\to x'^-}f'(x)$.

I'm a little uncomfortable with that, too, since it assumes a derivative exists over some ball about x', except for maybe at x' itself. So a better test would be one using the definition of derivative:
$f'(x) \equiv lim_{Δx\to0} \frac {f(x+Δx) - f(x)}{Δx}$
So if you can show
$lim_{Δx\to0^+} \frac {f(x+Δx) - f(x)}{Δx} = lim_{Δx\to0^-} \frac {f(x+Δx) - f(x)}{Δx}$
then $f(x)$ is differentiable at $x$.