Problem
No SIM 12:23 PM () $26 \%$ deltamath.com Score: 0/1 Penalty: 1 off Piecewise Differentiability with Transcenden Question Video Examples Given the piecewise function defined below, answer the questions at the bottom of the page regarding the continuity and differentiability of $f$ at $x=2$. \[ f(x)=\left\{\begin{array}{lll} 2 x^{2}-2 x & \text { for } & x<2 \\ 2 \ln (3 x-5)+4 & \text { for } & x \geq 2 \end{array}\right. \] Answer Attempt 1 out of 2 \[ \begin{array}{l} \lim _{x \rightarrow 2^{-}} f(x)=\square \quad f(2)=\square \\ \lim _{x \rightarrow 2^{+}} f(x)=\square \\ \lim _{x \rightarrow 2^{-}} f^{\prime}(x)=\square \\ \lim _{x \rightarrow 2^{+}} f^{\prime}(x)=\square \end{array} \] So the function $f(x)$ is at \[ x=2 \text {. } \] Submit Answer
Solution
Step 1 ：\[\lim_{x \rightarrow 2^-} f(x) = 2(2)^2 - 2(2) = 8 - 4 = 4\]
Step 2 ：\[f(2) = 2\ln(3(2) - 5) + 4 = 2\ln(1) + 4 = 4\]
Step 3 ：\[\lim_{x \rightarrow 2^+} f(x) = 2\ln(3(2) - 5) + 4 = 4\]
Step 4 ：\[\lim_{x \rightarrow 2^-} f'(x) = 4(2) - 2 = 6\]
Step 5 ：\[\lim_{x \rightarrow 2^+} f'(x) = 2/(3(2) - 5) = 2\]
Step 6 ：\[\text{The function f(x) is continuous at x=2 because the limit from the left, the limit from the right, and the value of the function at x=2 are all equal (4).}\]
Step 7 ：\[\text{However, the function is not differentiable at x=2 because the limit of the derivative from the left (6) does not equal the limit of the derivative from the right (2).}\]
