Question

Can I get help answering this question? It's a derivatives question, further info in the attachment.

Answer 1

`\bf f(x)=|3+9x|\implies f(x)=√((3+9x)^2)\implies f(x)=[(3+9x)^2]^{(1)/(2)} \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{2}[(3+9x)^2]^{-(1)/(2)}\cdot 2(3+9x)\cdot 9}\implies \cfrac{dy}{dx}=\cfrac{9(3+9x)}{[(3+9x)^2]^{(1)/(2)}} \\\\\\ \cfrac{dy}{dx}=\cfrac{27+81x}3+9x\\\\ -------------------------------`


`\bf \textit{left-hand derivative at }x=-(3)/(9)\implies \cfrac{27+81x}{-(3+9x)} \\\\\\ \cfrac{27+81\left( -(3)/(9) \right)}{-3-9\left( -(3)/(9) \right)}\implies \cfrac{27-27}{-3+3}\implies \stackrel{unde fined}{\cfrac{0}{0}}\\\\ -------------------------------\\\\ \textit{right-hand derivative at }x=-(3)/(9)\implies \cfrac{27+81x}{+(3+9x)} \\\\\\ \cfrac{27+81\left( -(3)/(9) \right)}{3+9\left( -(3)/(9) \right)}\implies \cfrac{27-27}{3-3}\implies \stackrel{unde fined}{\cfrac{0}{0}}`