Question

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

Answer 1

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