Find the critical point and the interval on which the given function is increasing or decreasing, and apply the First Derivative Test to the critical point. Let \( f(x)=3 \ln (4 x)-x, x>0 \) Critical Point \( = \) Is \( f \) a maximum or minumum at the critical point? ? The interval on the left of the critical point is On this interval, \( f \) is ? The interval on the right of the critical point is On this interval, \( f \) is ?

To find the critical point, we start by taking the derivative of the function \( f(x) = 3 \ln(4x) - x \). First, the derivative is: \[ f'(x) = \frac{3}{4x} \cdot 4 - 1 = \frac{3}{x} - 1 \] We set \( f'(x) = 0 \): \[ \frac{3}{x} - 1 = 0 \implies \frac{3}{x} = 1 \implies x = 3 \] **Critical Point**: \( x = 3 \) Now, we analyze the intervals around the critical point to see where the function is increasing or decreasing. 1. For the interval \( (0, 3) \): - Choose a test point, like \( x = 1 \): \[ f'(1) = \frac{3}{1} - 1 = 2 \quad ( > 0) \] Therefore, \( f \) is **increasing** on \( (0, 3) \). 2. For the interval \( (3, \infty) \): - Choose a test point, like \( x = 4 \): \[ f'(4) = \frac{3}{4} - 1 = -\frac{1}{4} \quad ( < 0) \] Therefore, \( f \) is **decreasing** on \( (3, \infty) \). **Applying the First Derivative Test**: Since \( f' \) changes from positive to negative at \( x = 3 \), we conclude that this critical point is a **maximum**. **Summary**: - **Critical Point**: \( x = 3 \) - On the left of the critical point \( (0, 3) \), \( f \) is **increasing**. - On the right of the critical point \( (3, \infty) \), \( f \) is **decreasing**. - \( f \) has a **maximum at \( x = 3 \)**.