Question:
If \( f(x) = \begin{cases} x^3 - 3x^2 - 12x + 1, & -1 \le x \le 2 \\ 37 - 2x, & 2 < x \le 3 \end{cases} \), then
If \( f(x) = \begin{cases} x^3 - 3x^2 - 12x + 1, & -1 \le x \le 2 \\ 37 - 2x, & 2 < x \le 3 \end{cases} \), then
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Differentiability implies continuity. If function is discontinuous at a point, derivative does not exist.
Updated On: May 2, 2026
- $f(x)$ is decreasing on $[-1,2]$
- $f'(2)$ does not exist
- $f(x)$ has the maximum value at $x = 2$
- None of the above
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Check continuity and differentiability at $x=2$.
Step 2: Detailed Explanation:
For $x \le 2$, $f(x) = x^3 - 3x^2 - 12x + 1$. At $x=2$, $f(2) = 8 - 12 - 24 + 1 = -27$.
For $x>2$, $f(x) = 37 - 2x$. At $x=2$, $f(2) = 37 - 4 = 33$.
Since $f(2)$ from left is $-27$ and from right is $33$, $f$ is not continuous at $x=2$. Hence $f'(2)$ does not exist.
Step 3: Final Answer:
$f'(2)$ does not exist.
Check continuity and differentiability at $x=2$.
Step 2: Detailed Explanation:
For $x \le 2$, $f(x) = x^3 - 3x^2 - 12x + 1$. At $x=2$, $f(2) = 8 - 12 - 24 + 1 = -27$.
For $x>2$, $f(x) = 37 - 2x$. At $x=2$, $f(2) = 37 - 4 = 33$.
Since $f(2)$ from left is $-27$ and from right is $33$, $f$ is not continuous at $x=2$. Hence $f'(2)$ does not exist.
Step 3: Final Answer:
$f'(2)$ does not exist.
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