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 ?

**Step 1. Compute the derivative of \( f(x) \):** \[ f(x)=3 \ln (4x)-x \] Differentiate with respect to \( x \): \[ f'(x)=3 \cdot \frac{1}{4x} \cdot 4-1=\frac{3}{x}-1 \] --- **Step 2. Find the critical point by setting \( f'(x)=0 \):** \[ \frac{3}{x}-1=0 \quad \Rightarrow \quad \frac{3}{x}=1 \quad \Rightarrow \quad x=3 \] The critical point is \( x=3 \). --- **Step 3. Determine the sign of \( f'(x) \) on intervals around the critical point:** - For \( x \) in the interval \( (0,3) \), choose \( x=1 \): \[ f'(1)=\frac{3}{1}-1=2>0 \quad \text{(positive)} \] - For \( x \) in the interval \( (3, \infty) \), choose \( x=4 \): \[ f'(4)=\frac{3}{4}-1=-\frac{1}{4}<0 \quad \text{(negative)} \] --- **Step 4. Apply the First Derivative Test:** Since \( f'(x) \) changes from positive to negative at \( x=3 \), the function \( f \) is increasing on the interval \( (0,3) \) and decreasing on the interval \( (3, \infty) \). Therefore, \( f \) has a local maximum at \( x=3 \). --- **Summary of Answers:** - **Critical Point:** \( x=3 \) - **Is \( f \) a maximum or minimum at the critical point?** Maximum - **The interval on the left of the critical point is:** \( (0,3) \) - **On this interval, \( f \) is:** Increasing - **The interval on the right of the critical point is:** \( (3, \infty) \) - **On this interval, \( f \) is:** Decreasing