if f(x) = `tan^(1) x -(1//2) ln x.` then

Options

A. the greatest value of f(x) on `[ 1//sqrt(3) , sqrt(3)]` is `pi//6 +(1//4) en 3`
B. the least value of f(x) on `[ 1//sqrt(3),sqrt(3) ] " is " pi//3 -(1//4) en 3`
C. f(x) decreases on `(0,oo)`
D. f(x) increases on `(-oo,0)`

Answer

To solve the problem step by step, we need to analyze the function given and find its critical points and behavior. The function is:

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

### Step 1: Find the derivative of \( f(x) \)

To find the critical points, we first need to compute the derivative \( f'(x) \):

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

Using the derivatives:
- The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \)
- The derivative of \( \ln(x) \) is \( \frac{1}{x} \)

Thus, we have:

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

### Step 2: Simplify the derivative

Now, we need to simplify \( f'(x) \):

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

To combine these fractions, we find a common denominator:

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

Rearranging gives:

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

### Step 3: Find critical points

Setting \( f'(x) = 0 \):

\[ -(x-1)^2 = 0 \]

This gives us the critical point:

\[ x = 1 \]

### Step 4: Analyze the sign of \( f'(x) \)

To determine the behavior of \( f(x) \), we analyze the sign of \( f'(x) \):

- For \( x < 1 \), \( (x-1)^2 > 0 \) hence \( f'(x) < 0 \) (decreasing).
- For \( x > 1 \), \( (x-1)^2 > 0 \) hence \( f'(x) < 0 \) (decreasing).

Thus, \( f(x) \) is decreasing for all \( x > 0 \).

### Step 5: Evaluate \( f(x) \) at the endpoints

Now we need to evaluate \( f(x) \) at the endpoints of the interval \( \left[ \frac{1}{\sqrt{3}}, \sqrt{3} \right] \):

1. **At \( x = \frac{1}{\sqrt{3}} \)**:
\[ f\left(\frac{1}{\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \frac{1}{2} \ln\left(\frac{1}{\sqrt{3}}\right) \]
\[ = \frac{\pi}{6} + \frac{1}{2} \ln(\sqrt{3}) = \frac{\pi}{6} + \frac{1}{4} \ln(3) \]

2. **At \( x = \sqrt{3} \)**:
\[ f(\sqrt{3}) = \tan^{-1}(\sqrt{3}) - \frac{1}{2} \ln(\sqrt{3}) \]
\[ = \frac{\pi}{3} - \frac{1}{4} \ln(3) \]

### Step 6: Compare values

Since \( f(x) \) is decreasing, the greatest value of \( f(x) \) on the interval \( \left[ \frac{1}{\sqrt{3}}, \sqrt{3} \right] \) occurs at \( x = \frac{1}{\sqrt{3}} \).

### Conclusion

The greatest value of \( f(x) \) on the interval \( \left[ \frac{1}{\sqrt{3}}, \sqrt{3} \right] \) is:

\[ f\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} + \frac{1}{4} \ln(3) \]