Question

If \( M \) and \( m \) denote the local maximum and local minimum values of the function \( f(x) = x + \frac{1}{x} \) (\( x \neq 0 \)) respectively, find the value of \( M - m \).

Hint:

To find local maxima and minima, first use the first derivative to locate critical points, and then use the second derivative test to classify them.

Answer 1

Step 1: Define the function and differentiate.
The given function is: \[ f(x) = x + \frac{1}{x}. \] To find critical points, we differentiate \( f(x) \): \[ f'(x) = 1 - \frac{1}{x^2}. \] Step 2: Find the critical points.
Set the derivative equal to zero to solve for critical points: \[ f'(x) = 0 \quad \Rightarrow \quad 1 - \frac{1}{x^2} = 0 \quad \Rightarrow \quad \frac{1}{x^2} = 1 \quad \Rightarrow \quad x^2 = 1. \] This gives two critical points: \[ x = 1 \quad \text{or} \quad x = -1. \] Step 3: Classify the critical points.
To determine the nature of these critical points, we compute the second derivative of \( f(x) \): \[ f''(x) = \frac{2}{x^3}. \] - At \( x = 1 \): \[ f''(1) = \frac{2}{1^3} = 2 \quad (\text{positive, so } x = 1 \text{ is a local minimum}). \] - At \( x = -1 \): \[ f''(-1) = \frac{2}{(-1)^3} = -2 \quad (\text{negative, so } x = -1 \text{ is a local maximum}). \] Step 4: Evaluate the function at the critical points.
- At \( x = 1 \): \[ f(1) = 1 + \frac{1}{1} = 2 \quad (\text{local minimum } m). \] - At \( x = -1 \): \[ f(-1) = -1 + \frac{1}{-1} = -2 \quad (\text{local maximum } M). \] Step 5: Calculate \( M - m \).
Now, calculate the difference between the maximum and minimum values: \[ M - m = -2 - 2 = -4. \] Conclusion:
Thus, the value of \( M - m \) is: \[ \boxed{-4}. \]