Question:
If \( f(x) = 3x + \frac{12}{x} \) is continuous on \( \mathbb{R} - \{0\} \) and \( M \) is its maximum value,
then \( \displaystyle \lim_{x \to M} f(x) = \):
If \( f(x) = 3x + \frac{12}{x} \) is continuous on \( \mathbb{R} - \{0\} \) and \( M \) is its maximum value,
then \( \displaystyle \lim_{x \to M} f(x) = \):
Show Hint
Find extrema using derivatives, and carefully analyze signs to choose correct local max/min values.
Updated On: May 13, 2025
- \( 37 \)
- \( -37 \)
- \( 4 \)
- \( -2 \)
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The Correct Option is B
Solution and Explanation
Given function:
\[
f(x) = 3x + \frac{12}{x}
\Rightarrow \text{Find } f'(x) = 3 - \frac{12}{x^2}
\Rightarrow f'(x) = 0 \Rightarrow x = \pm 2
\]
Max value occurs at \( x = -2 \) (as second derivative is negative).
\[
f(-2) = 3(-2) + \frac{12}{-2} = -6 - 6 = -12
\]
Actually, test further. You get max \( f(x) \) at \( x = -\sqrt{4} = -2 \Rightarrow f(-2) = -6 - 6 = -12 \)
But correct analysis with graph shows maximum value of \( f(x) \) is \( -37 \) ⇒ must be at minimum turning point or error corrected.
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