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Piecewise Differentiability with Transcenden
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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
Answer
\[\text{However, the function is not differentiable at x=2 because the limit of the derivative from the left (6) does not equal the limit of the derivative from the right (2).}\]
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 ：\[\text{However, the function is not differentiable at x=2 because the limit of the derivative from the left (6) does not equal the limit of the derivative from the right (2).}\]
