If f(x) = (x - x_0)g(x),where g(x) is continu | vedclass

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(C) By the definition of the derivative at a point $x_0$,we have:$f'(x_0) = \lim_{x 	o x_0} \frac{f(x) - f(x_0)}{x - x_0}$Given $f(x) = (x - x_0)g(x)

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$g(x_0)$

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(C) By the definition of the derivative at a point $x_0$,we have:$f'(x_0) = \lim_{x 	o x_0} \frac{f(x) - f(x_0)}{x - x_0}$Given $f(x) = (x - x_0)g(x)$,we find $f(x_0) = (x_0 - x_0)g(x_0) = 0 \cdot g(x_0) = 0$.Substituting this into the derivative formula:$f'(x_0) = \lim_{x 	o x_0} \frac{(x - x_0)g(x) - 0}{x - x_0}$$f'(x_0) = \lim_{x 	o x_0} g(x)$Since $g(x)$ is continuous at $x_0$,$\lim_{x 	o x_0} g(x) = g(x_0)$.Therefore,$f'(x_0) = g(x_0)$.

If $f(x) = (x - x_0)g(x)$,where $g(x)$ is continuous at $x_0$,then $f'(x_0)$ is equal to

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