If f(x) = `log_(x) 1//9 - log_(3) x^(2) (x gt 1)` then |max f(x)| is equal to

Answer

To solve the problem, we need to find the maximum value of the function \( f(x) = \log_x \left( \frac{1}{9} \right) - \log_3 (x^2) \) for \( x > 1 \).

### Step-by-Step Solution:

1. **Rewrite the logarithmic terms**:
We can express \( \frac{1}{9} \) as \( 9^{-1} \) and use the properties of logarithms:
\[ f(x) = \log_x(9^{-1}) - \log_3(x^2) \]
Using the property \( \log_a(b^c) = c \cdot \log_a(b) \), we can rewrite:
\[ f(x) = -\log_x(9) - 2\log_3(x) \]

2. **Change of base for the first term**:
We can convert \( \log_x(9) \) using the change of base formula:
\[ \log_x(9) = \frac{\log_3(9)}{\log_3(x)} = \frac{2}{\log_3(x)} \]
Therefore, we have:
\[ f(x) = -\frac{2}{\log_3(x)} - 2\log_3(x) \]

3. **Let \( y = \log_3(x) \)**:
Since \( x > 1 \), \( y > 0 \). Then we can rewrite the function:
\[ f(y) = -\frac{2}{y} - 2y \]

4. **Find the derivative**:
To find the maximum, we take the derivative of \( f(y) \):
\[ f'(y) = \frac{2}{y^2} - 2 \]
Set the derivative to zero to find critical points:
\[ \frac{2}{y^2} - 2 = 0 \]
\[ \frac{2}{y^2} = 2 \implies y^2 = 1 \implies y = 1 \quad (\text{since } y > 0) \]

5. **Evaluate the second derivative**:
To confirm that this is a maximum, we check the second derivative:
\[ f''(y) = -\frac{4}{y^3} \]
Since \( f''(y) < 0 \) for \( y > 0 \), this indicates a local maximum at \( y = 1 \).

6. **Calculate the maximum value**:
Substitute \( y = 1 \) back into \( f(y) \):
\[ f(1) = -\frac{2}{1} - 2(1) = -2 - 2 = -4 \]

7. **Find the absolute value**:
Since we need \( |max f(x)| \):
\[ |max f(x)| = |-4| = 4 \]

### Final Answer:
The maximum value of \( |f(x)| \) is \( 4 \).