Let `f(x) =x +(1)/(x).` find local maximum and local minimum value of `f(x) .` Can you explain this discrepancy of locally minimum value being greater than locally maximum value.

Answer

To find the local maximum and minimum values of the function \( f(x) = x + \frac{1}{x} \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \) with respect to \( x \).

\[ f'(x) = 1 - \frac{1}{x^2} \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:

\[ 1 - \frac{1}{x^2} = 0 \]

This simplifies to:

\[ \frac{1}{x^2} = 1 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1 \]

### Step 3: Determine the nature of the critical points
Next, we will use the second derivative test to determine whether these critical points are local maxima or minima. First, we find the second derivative:

\[ f''(x) = \frac{2}{x^3} \]

Now we evaluate the second derivative at the critical points:

1. For \( x = 1 \):
\[ f''(1) = \frac{2}{1^3} = 2 > 0 \quad \text{(local minimum)} \]

2. For \( x = -1 \):
\[ f''(-1) = \frac{2}{(-1)^3} = -2 < 0 \quad \text{(local maximum)} \]

### Step 4: Calculate the local maximum and minimum values
Now we will calculate the function values at these critical points:

1. For \( x = 1 \):
\[ f(1) = 1 + \frac{1}{1} = 2 \]

2. For \( x = -1 \):
\[ f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2 \]

### Conclusion
Thus, we have:
- Local maximum at \( x = -1 \) with \( f(-1) = -2 \)
- Local minimum at \( x = 1 \) with \( f(1) = 2 \)

### Explanation of the discrepancy
The discrepancy arises because the local minimum value (2) is greater than the local maximum value (-2). This is due to the nature of the function \( f(x) \) which approaches infinity as \( x \) approaches 0 from either side and has a minimum value at \( x = 1 \) that is higher than the maximum value at \( x = -1 \). The function is not bounded below, leading to this situation.