hi, can someone solve this? ​

Answer:Possible interpretations and solutions:  Evaluate \(f(x)\) at specific points: . To evaluate the function, determine which interval the input \(x\) falls into and use the corresponding sub-function. For example, to find \(f(-1)\): Since \(-1<0\), use the first rule: \(f(-1)=2(-1)-2=-2-2=-4\). To find \(f(1)\): Since \(0\le 1<3\), use the second rule: \(f(1)=-1\). To find \(f(3)\): Since \(3\ge 3\), use the third rule: \(f(3)=-4(3)+16=-12+16=4\). 2. Determine if \(f(x)\) is continuous at specific points (e.g., \(x=0\) and \(x=3\)): Continuity at \(x=0\): Left-hand limit: \(\lim _{x\rightarrow 0^{-}}f(x)=\lim _{x\rightarrow 0^{-}}(2x-2)=2(0)-2=-2\). Right-hand limit: \(\lim _{x\rightarrow 0^{+}}f(x)=\lim _{x\rightarrow 0^{+}}(-1)=-1\). Since the left-hand limit (\(-2\)) does not equal the right-hand limit (\(-1\)), the function is not continuous at \(x=0\). Continuity at \(x=3\): Left-hand limit: \(\lim _{x\rightarrow 3^{-}}f(x)=\lim _{x\rightarrow 3^{-}}(-1)=-1\). Right-hand limit: \(\lim _{x\rightarrow 3^{+}}f(x)=\lim _{x\rightarrow 3^{+}}(-4x+16)=-4(3)+16=-12+16=4\). Since the left-hand limit (\(-1\)) does not equal the right-hand limit (\(4\)), the function is not continuous at \(x=3\).

Answered by norielmanikan | 2025-07-27