If f(x) = begin{cases} frac{x^2-16}{x-4} & te | vedclass

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(C) Given,$f(x) = \begin{cases} \frac{x^2-16}{x-4} & 	ext{if } x > 4 \ 2x & 	ext{if } x \leq 4 \end{cases}$ For $x > 4$,$f(x) = \frac{(x-4)(x+4)}{x

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$3$

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(C) Given,$f(x) = \begin{cases} \frac{x^2-16}{x-4} & 	ext{if } x > 4 \ 2x & 	ext{if } x \leq 4 \end{cases}$ For $x > 4$,$f(x) = \frac{(x-4)(x+4)}{x-4} = x+4$. For $x \leq 4$,$f(x) = 2x$. Now,differentiating with respect to $x$: $f^{\prime}(x) = \begin{cases} \frac{d}{dx}(x+4) = 1 & 	ext{if } x > 4 \ \frac{d}{dx}(2x) = 2 & 	ext{if } x Therefore,$f^{\prime}(4^{+}) = \lim_{h 	o 0} f^{\prime}(4+h) = 1$ and $f^{\prime}(4^{-}) = \lim_{h 	o 0} f^{\prime}(4-h) = 2$. Thus,$f^{\prime}(4^{-}) + f^{\prime}(4^{+}) = 2 + 1 = 3$.

If $f(x) = \begin{cases} \frac{x^2-16}{x-4} & \text{if } x > 4 \\ 2x & \text{if } x \leq 4 \end{cases}$ then $f^{\prime}(4^{-}) + f^{\prime}(4^{+}) = $

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