The maximum & minimum value of `y=x+(1)/(x)` in interval `[(1)/(3),(4)/(3)]`

Options

A. 2,-2
B. `(10)/(3),2`
C. `oo,-oo`
D. None of these

Answer

To find the maximum and minimum values of the function \( y = x + \frac{1}{x} \) in the interval \( \left[\frac{1}{3}, \frac{4}{3}\right] \), we will follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function with respect to \( x \):
\[ \frac{dy}{dx} = 1 - \frac{1}{x^2} \]

### Step 2: Set the derivative to zero
To find the critical points where the function may attain maximum or minimum values, we set the derivative equal to zero:
\[ 1 - \frac{1}{x^2} = 0 \]

### Step 3: Solve for \( x \)
Rearranging the equation gives:
\[ \frac{1}{x^2} = 1 \implies x^2 = 1 \implies x = \pm 1 \]
Since we are looking for values in the interval \( \left[\frac{1}{3}, \frac{4}{3}\right] \), we only consider \( x = 1 \).

### Step 4: Evaluate the function at the critical point and endpoints
Next, we need to evaluate the function \( y \) at the critical point \( x = 1 \) and at the endpoints of the interval \( x = \frac{1}{3} \) and \( x = \frac{4}{3} \).

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

2. For \( x = \frac{1}{3} \):
\[ y\left(\frac{1}{3}\right) = \frac{1}{3} + \frac{1}{\frac{1}{3}} = \frac{1}{3} + 3 = \frac{10}{3} \]

3. For \( x = \frac{4}{3} \):
\[ y\left(\frac{4}{3}\right) = \frac{4}{3} + \frac{1}{\frac{4}{3}} = \frac{4}{3} + \frac{3}{4} = \frac{16}{12} + \frac{9}{12} = \frac{25}{12} \]

### Step 5: Compare the values
Now we compare the values obtained:
- \( y(1) = 2 \)
- \( y\left(\frac{1}{3}\right) = \frac{10}{3} \approx 3.33 \)
- \( y\left(\frac{4}{3}\right) = \frac{25}{12} \approx 2.08 \)

### Conclusion
The minimum value of \( y \) in the interval is \( 2 \) at \( x = 1 \), and the maximum value is \( \frac{10}{3} \) at \( x = \frac{1}{3} \).

### Final Answer
- Minimum value: \( 2 \)
- Maximum value: \( \frac{10}{3} \)