If `f(x)=x+(1)/(x),xne0`, then local maximum and minimum values of function f are respectively . . .

Options

A. `-1 and 1`
B. `-2 and 2`
C. `2 and -2`
D. `1 and -1`

Explanation

To find the local maximum and minimum values of the function \( f(x) = x + \frac{1}{x} \) where \( x \neq 0 \), we will follow these steps:

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

\[ f'(x) = \frac{d}{dx} \left( x + \frac{1}{x} \right) = 1 - \frac{1}{x^2} \]

**Hint:** Remember that the derivative gives us the slope of the function, which helps us find critical points.

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

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

**Hint:** Setting the derivative to zero helps us find points where the function may have a local maximum or minimum.

### Step 3: Solve for \( x \)
Rearranging the equation gives:

\[ \frac{1}{x^2} = 1 \implies x^2 = 1 \implies x = \pm 1 \]

**Hint:** When solving for \( x \), consider both the positive and negative roots.

### Step 4: Determine the nature of the critical points
Next, we need to determine whether these critical points are local maxima or minima. We can use the first derivative test.

1. For \( x = -1 \):
- Choose a test point in the interval \( (-\infty, -1) \), say \( x = -2 \):
\[ f'(-2) = 1 - \frac{1}{(-2)^2} = 1 - \frac{1}{4} = \frac{3}{4} > 0 \quad (\text{increasing}) \]
- Choose a test point in the interval \( (-1, 0) \), say \( x = -0.5 \):
\[ f'(-0.5) = 1 - \frac{1}{(-0.5)^2} = 1 - 4 = -3 < 0 \quad (\text{decreasing}) \]

Since \( f'(-2) > 0 \) and \( f'(-0.5) < 0 \), we conclude that \( x = -1 \) is a local maximum.

2. For \( x = 1 \):
- Choose a test point in the interval \( (0, 1) \), say \( x = 0.5 \):
\[ f'(0.5) = 1 - \frac{1}{(0.5)^2} = 1 - 4 = -3 < 0 \quad (\text{decreasing}) \]
- Choose a test point in the interval \( (1, \infty) \), say \( x = 2 \):
\[ f'(2) = 1 - \frac{1}{(2)^2} = 1 - \frac{1}{4} = \frac{3}{4} > 0 \quad (\text{increasing}) \]

Since \( f'(0.5) < 0 \) and \( f'(2) > 0 \), we conclude that \( x = 1 \) is a local minimum.

**Hint:** The first derivative test helps us determine whether the function is increasing or decreasing around the critical points.

### Step 5: Calculate the local maximum and minimum values
Now we can find the values of \( f(x) \) at these critical points:

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

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

### Conclusion
Thus, the local maximum value is \( -2 \) at \( x = -1 \) and the local minimum value is \( 2 \) at \( x = 1 \).

**Final Answer: Local Maximum is -2, Local Minimum is 2.**