Question

For the function $f(x) = x + \frac{1}{x}$ ($x \neq 0$), which of the following statements is true?

• 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:

For $x > 0$, by the AM-GM inequality: $\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = 1 \Rightarrow x + \frac{1}{x} \geq 2$. Thus, the minimum positive value is 2. For negative values, replacing $x$ with $-x$ flips the signs, making the maximum negative value $-2$.

Answer 1

The Correct Option is C

Concept: To find local extrema (maxima and minima) of a continuous differentiable function $f(x)$, we apply the first and second derivative tests:
• Find critical points where $f'(x) = 0$.
• Evaluate $f''(x)$ at these points. If $f''(x) > 0$, it represents a local minimum. If $f''(x) < 0$, it represents a local maximum.

Step 1: Compute the first derivative and find critical points.
The given function is: \[ f(x) = x + x^{-1} \] Differentiating with respect to $x$: \[ f'(x) = 1 - \frac{1}{x^2} \] To find the critical points, set $f'(x) = 0$: \[ 1 - \frac{1}{x^2} = 0 \quad \Rightarrow \quad 1 = \frac{1}{x^2} \quad \Rightarrow \quad x^2 = 1 \] Taking the square root of both sides gives two critical points: \[ x = 1 \quad \text{and} \quad x = -1 \]

Step 2: Compute the second derivative to test the nature of critical points.
Now, differentiate $f'(x) = 1 - x^{-2}$ again with respect to $x$: \[ f''(x) = 0 - (-2)x^{-3} = \frac{2}{x^3} \] Let us test each critical point separately:
• At $x = 1$: \[ f''(1) = \frac{2}{(1)^3} = 2 > 0 \] Since the second derivative is positive, $x = 1$ is a point of local minimum.
• At $x = -1$: \[ f''(-1) = \frac{2}{(-1)^3} = -2 < 0 \] Since the second derivative is negative, $x = -1$ is a point of local maximum.

Step 3: Calculate the local maximum and local minimum values.

• Local Minimum Value occurs at $x = 1$: \[ f(1) = 1 + \frac{1}{1} = 2
• Local Maximum Value occurs at $x = -1$: \[ f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2 Looking at the options, option (C) states that the local maximum value is $-2$, which matches our findings perfectly. Option (D) is false because the local minimum value ($2$) is actually greater than the local maximum value ($-2$).