The left-hand derivative of f(x) = [x]sin(pi | vedclass

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(A) The left-hand derivative at $x = k$ is defined as $f'(k^-) = \lim_{h 	o 0^+} \frac{f(k - h) - f(k)}{-h}$.Given $f(x) = [x]\sin(\pi x)$,we have $f

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$(-1)^k(k - 1)\pi$

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(A) The left-hand derivative at $x = k$ is defined as $f'(k^-) = \lim_{h 	o 0^+} \frac{f(k - h) - f(k)}{-h}$.Given $f(x) = [x]\sin(\pi x)$,we have $f(k) = [k]\sin(\pi k) = k 	imes 0 = 0$.For small $h > 0$,$[k - h] = k - 1$.Thus,$f(k - h) = (k - 1)\sin(\pi(k - h)) = (k - 1)\sin(\pi k - \pi h) = (k - 1)(\sin(\pi k)\cos(\pi h) - \cos(\pi k)\sin(\pi h))$.Since $\sin(\pi k) = 0$ and $\cos(\pi k) = (-1)^k$,we get $f(k - h) = (k - 1)(0 - (-1)^k \sin(\pi h)) = -(k - 1)(-1)^k \sin(\pi h) = (k - 1)(-1)^{k+1} \sin(\pi h)$.Substituting this into the limit: $f'(k^-) = \lim_{h 	o 0^+} \frac{(k - 1)(-1)^{k+1} \sin(\pi h) - 0}{-h}$.Using $\lim_{h 	o 0} \frac{\sin(\pi h)}{\pi h} = 1$,we get $f'(k^-) = (k - 1)(-1)^{k+1} 	imes (-\pi) = (k - 1)(-1)^{k+1} (-1) \pi = (k - 1)(-1)^{k+2} \pi = (k - 1)(-1)^k \pi$.

The left-hand derivative of $f(x) = [x]\sin(\pi x)$ at $x = k$,where $k$ is an integer and $[x]$ denotes the greatest integer function $\le x$,is:

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