If f(x) = begin{cases} x, & 0 le x le 1 2x - | vedclass

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(C) Given $f(x) = \begin{cases} x, & 0 \le x \le 1 \ 2x - 1, & x > 1 \end{cases}$.First,we check for continuity at $x = 1$:$\lim_{x 	o 1^-} f(x) = \

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$f$ is continuous but not differentiable at $x = 1$

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(C) Given $f(x) = \begin{cases} x, & 0 \le x \le 1 \ 2x - 1, & x > 1 \end{cases}$.First,we check for continuity at $x = 1$:$\lim_{x 	o 1^-} f(x) = \lim_{h 	o 0} f(1 - h) = \lim_{h 	o 0} (1 - h) = 1$.$\lim_{x 	o 1^+} f(x) = \lim_{h 	o 0} f(1 + h) = \lim_{h 	o 0} (2(1 + h) - 1) = \lim_{h 	o 0} (2 + 2h - 1) = 1$.Since $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1) = 1$,the function is continuous at $x = 1$.Next,we check for differentiability at $x = 1$:Left-hand derivative: $Lf'(1) = \lim_{h 	o 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h 	o 0} \frac{(1 - h) - 1}{-h} = \lim_{h 	o 0} \frac{-h}{-h} = 1$.Right-hand derivative: $Rf'(1) = \lim_{h 	o 0} \frac{f(1 + h) - f(1)}{h} = \lim_{h 	o 0} \frac{(2(1 + h) - 1) - 1}{h} = \lim_{h 	o 0} \frac{2 + 2h - 2}{h} = \lim_{h 	o 0} \frac{2h}{h} = 2$.Since $Lf'(1) 
eq Rf'(1)$,the function is not differentiable at $x = 1$.Thus,$f$ is continuous but not differentiable at $x = 1$.

If $f(x) = \begin{cases} x, & 0 \le x \le 1 \\ 2x - 1, & x > 1 \end{cases}$,then

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