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In which points is the following function differentiable (in non-distributional sense)?

$$ f(t)= t \theta(t) + (1/2t^2-t+1/2)\theta(t-1)+(t-2)\theta(t-2)$$

My solution:

\begin{align*} f'(t)&= \theta(t) + (t-1) \theta(t-1)+\theta(t-2)\\ f''(t)&= \delta(t) + \theta(t-1)+\delta(t-2) \end{align*}

However, I am not sure how I should interpret my results. A part of belives every point except $t=0,1,2$ when looking at $f(t)$. However when looking at $f'(t)$ and $f''(t)$ I think that it might be every point except $t=0,2$.

How should I interpret my $f'$ and $f''$ and will it help to solve my question?

cmk
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JavaMan
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  • The derivative is only contious in every point but 0,1,2 or is the derivative only contious in every point except 0 and 2? – JavaMan Jun 01 '19 at 19:45
  • It is a sum of piecewise smooth functions and the sum is continuous, thus it is the primitive of a piecewise smooth function and it is differentiable everywhere the derivative is continuous (at the discontinuity points $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists only as one-sided limits). $f'(t)$ is continuous at $t=1$ so $\int_0^t f'(u)du$ is differentiable at $t=1$. The point is that $f'(t)$ has a jump discontinuity at $t=0$ thus $\int_0^t f'(u)du$ is not differentiable at $t=0$ (try with $g(t) = |t|$) – reuns Jun 01 '19 at 20:07

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