The least value of the function `F(x) = int_(x)^(2) log_(1//3) t dt, x in [1//10,4]` is at `x=`

Options

A. `1//10`
B. `4`
C. `1`
D. none of these

Answer

To find the least value of the function \( F(x) = \int_{x}^{2} \log_{1/3} t \, dt \) for \( x \in \left[\frac{1}{10}, 4\right] \), we will follow these steps:

### Step 1: Differentiate \( F(x) \)
Using the Leibniz rule for differentiation under the integral sign, we differentiate \( F(x) \):

\[ F'(x) = -\log_{1/3} x \]

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

\[ -\log_{1/3} x = 0 \]

This implies:

\[ \log_{1/3} x = 0 \]

### Step 3: Solve for \( x \)
The equation \( \log_{1/3} x = 0 \) means:

\[ x = 3^0 = 1 \]

### Step 4: Check the critical point within the interval
We need to check if \( x = 1 \) lies within the interval \( \left[\frac{1}{10}, 4\right] \). Since \( 1 \) is indeed within this interval, we will evaluate \( F(x) \) at the endpoints and at the critical point.

### Step 5: Evaluate \( F(x) \) at the endpoints and the critical point
1. **At \( x = \frac{1}{10} \)**:
\[ F\left(\frac{1}{10}\right) = \int_{\frac{1}{10}}^{2} \log_{1/3} t \, dt \]

2. **At \( x = 1 \)**:
\[ F(1) = \int_{1}^{2} \log_{1/3} t \, dt \]

3. **At \( x = 4 \)**:
\[ F(4) = \int_{4}^{2} \log_{1/3} t \, dt = -\int_{2}^{4} \log_{1/3} t \, dt \]

### Step 6: Compare the values
We will need to calculate these integrals to find the least value. However, we can analyze the behavior of \( F(x) \) based on the sign of \( F'(x) \):

- For \( x < 1 \), \( F'(x) > 0 \) (function is increasing).
- For \( x > 1 \), \( F'(x) < 0 \) (function is decreasing).

Thus, \( F(x) \) attains its minimum at \( x = 1 \).

### Conclusion
The least value of the function \( F(x) \) occurs at:

\[ \boxed{1} \]