f(x) = begin{cases} [cos pi x]; & x leqslant | vedclass

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(D) First,we find the value of the function at $x = 1$: $f(1) = [\cos(\pi)] = [-1] = -1$.Now,we calculate the left-hand derivative at $x = 1$:$f'(1^-)

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All of the above

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(D) First,we find the value of the function at $x = 1$: $f(1) = [\cos(\pi)] = [-1] = -1$.Now,we calculate the left-hand derivative at $x = 1$:$f'(1^-) = \lim_{h 	o 0^+} \frac{f(1-h) - f(1)}{-h} = \lim_{h 	o 0^+} \frac{[\cos(\pi(1-h))] - (-1)}{-h} = \lim_{h 	o 0^+} \frac{[\cos(\pi - \pi h)] + 1}{-h} = \lim_{h 	o 0^+} \frac{[-\cos(\pi h)] + 1}{-h}$.Since for small $h > 0$,$\cos(\pi h)$ is slightly less than $1$,$-\cos(\pi h)$ is slightly greater than $-1$. Thus,$[-\cos(\pi h)] = -1$.So,$f'(1^-) = \lim_{h 	o 0^+} \frac{-1 + 1}{-h} = 0$.Next,we calculate the right-hand derivative at $x = 1$:$f'(1^+) = \lim_{h 	o 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0^+} \frac{2\{1+h\} - 1 - (-1)}{h} = \lim_{h 	o 0^+} \frac{2h}{h} = 2$.Since $f'(1^-) = 0$ and $f'(1^+) = 2$,the left-hand derivative is not equal to the right-hand derivative.Therefore,$f(x)$ is not differentiable at $x = 1$.

$f(x) = \begin{cases} [\cos \pi x]; & x \leqslant 1 \\ 2\{x\} - 1; & x > 1 \end{cases}$ Comment on the derivability at $x = 1$,where $[\cdot]$ denotes the greatest integer function and $\{\cdot\}$ denotes the fractional part function.

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