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Calculate f '(0) for the following continuous function. (If an answer does not exist, enter DNE.)

f(x) = |x| + 4




f '(0) =

c

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\bf f(x)=|x|+4\implies f(x)=√(x^2)+4\implies f(x)=(x^2)^{(1)/(2)}+4 \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{\underline{2}}(x^2)^{-(1)/(2)}\cdot \underline{2} x}\implies \cfrac{dy}{dx}=\cfrac{x}{(x^2)^{(1)/(2)}} \\\\\\ \left. \cfrac{dy}{dx}=\cfrac{x}{√(x^2)} \right|_(x=0)\implies \stackrel{und efined}{\cfrac{0}{0}}
User Adam Rezich
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