Suppose f(x) = begin{cases} [cos pi x], & x l | vedclass

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(A) Given $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \ 2\{x\} - 1, & x > 1 \end{cases}$.For $x > 1$, we have $f(x) = 2(x - [x]) - 1$. Since for $

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right derivative is $2$

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(A) Given $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \ 2\{x\} - 1, & x > 1 \end{cases}$.For $x > 1$, we have $f(x) = 2(x - [x]) - 1$. Since for $x$ slightly greater than $1$, $[x] = 1$, we get $f(x) = 2(x - 1) - 1 = 2x - 3$.The right-hand derivative at $x = 1$ is defined as $f'(1^+) = \lim_{h 	o 0^+} \frac{f(1+h) - f(1)}{h}$.First, $f(1) = [\cos \pi] = [-1] = -1$.Then, $f'(1^+) = \lim_{h 	o 0^+} \frac{2(1+h) - 3 - (-1)}{h} = \lim_{h 	o 0^+} \frac{2 + 2h - 3 + 1}{h} = \lim_{h 	o 0^+} \frac{2h}{h} = 2$.Thus, the right derivative at $x = 1$ is $2$.

Suppose $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \\ 2\{x\} - 1, & x > 1 \end{cases}$, where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer function and the fractional part of $x$ respectively, then at $x = 1$:

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