Question

If \( f(x) = 3x + \frac{12}{x} \) is continuous on \( \mathbb{R} - \{0\} \) and \( M \) is its maximum value, then \( \displaystyle \lim_{x \to M} f(x) = \):

• A. \( 37 \)
• B. \( -37 \)
• C. \( 4 \)
• D. \( -2 \)

Hint:

Find extrema using derivatives, and carefully analyze signs to choose correct local max/min values.

Answer 1

The Correct Option is B

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.