Question

For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

• A. local maximum value is 2
• B. local minimum value is \( -2 \)
• C. local maximum value is \( -2 \)
• D. local minimum value \( < \) local maximum value

Hint:

Note that for this function, the local minimum value (2) is actually greater than the local maximum value (-2). This is common in functions with asymptotes.
Recall the AM-GM inequality: for \( x > 0 \), \( x + \frac{1}{x} \geq 2 \). This confirms the minimum at \( x = 1 \).

Answer 1

The Correct Option is C

Concept:
• Critical points occur where the first derivative \( f'(x) = 0 \).
• Second derivative test: If \( f'(c) = 0 \), then \( f''(c) > 0 \) indicates a local minimum and \( f''(c) < 0 \) indicates a local maximum.

Step 1: Find the first derivative and critical points
Given \( f(x) = x + x^{-1} \).
Differentiating:
\[ f'(x) = 1 - \frac{1}{x^2} \]
Set \( f'(x) = 0 \):
\[ 1 - \frac{1}{x^2} = 0 \implies x^2 = 1 \implies x = 1, -1 \]

Step 2: Find the second derivative
Differentiating \( f'(x) \):
\[ f''(x) = \frac{d}{dx}(1 - x^{-2}) = 2x^{-3} = \frac{2}{x^3} \]

Step 3: Test each critical point
For \( x = 1 \):
\[ f''(1) = \frac{2}{1^3} = 2 > 0 \]
Thus, \( x = 1 \) is a point of local minimum. Local minimum value \( f(1) = 1 + \frac{1}{1} = 2 \).
For \( x = -1 \):
\[ f''(-1) = \frac{2}{(-1)^3} = -2 < 0 \]
Thus, \( x = -1 \) is a point of local maximum. Local maximum value \( f(-1) = -1 + \frac{1}{-1} = -2 \).

Step 4: Compare results with options
Option (A): Local max is 2 (False, it is -2).
Option (B): Local min is -2 (False, it is 2).
Option (C): Local max is -2 (True).
Option (D): Local min (2) \( < \) Local max (-2) (False).