Let's analyze the graph step by step. We are given the graph of \( y = \int_0^x f(t) dt \), where \( f \) is a piecewise constant function. The graph shows the values of the integral function, not \( f(x) \) itself. ### (a) Over which intervals is \( f \) positive, negative, or zero? Recall: - If \( \frac{d}{dx}\left(\int_0^x f(t)dt\right) = f(x) \). - Where the graph is **increasing**, \( f(x) > 0 \). - Where the graph is **decreasing**, \( f(x) < 0 \). - Where the graph is **constant** (horizontal), \( f(x) = 0 \). #### 1. Where is \( f \) positive? Look for intervals where \( y \) increases: - From \( x = 0 \) to \( x = 1 \) (**increasing**) - From \( x = 2 \) to \( x = 4 \) (**increasing**) **Answer:** \( (0,1) \cup (2,4) \) #### 2. Where is \( f \) negative? Look for intervals where \( y \) decreases: - From \( x = 1 \) to \( x = 2 \) (**decreasing**) - From \( x = 4 \) to \( x = 6 \) (**decreasing**) **Answer:** \( (1,2) \cup (4,6) \) #### 3. Where is \( f \) zero? Look for intervals where \( y \) is constant (horizontal): - From \( x = 1 \) to \( x = 2 \) (not horizontal) - From \( x = 2 \) to \( x = 4 \) (not horizontal) - From \( x = 4 \) to \( x = 6 \) (not horizontal) - From \( x = 0 \) to \( x = 1 \) (not horizontal) - At \( x = 1, 2, 4 \) (the points where the slope changes) Let's check the graph carefully: - The graph is **horizontal** at \( x = 1, 2, 4 \) (these are the points, not intervals). So the correct answer is **at the points** \( x = 1, 2, 4 \). **Answer:** \( \{1, 2, 4\} \) --- ### (b) Maximum and minimum values of \( f \): Since \( f(x) \) is constant on each interval: - From \( x = 0 \) to \( x = 1 \), the slope is **2**. - From \( x = 1 \) to \( x = 2 \), the slope is **-3**. - From \( x = 2 \) to \( x = 4 \), the slope is **1**. - From \( x = 4 \) to \( x = 6 \), the slope is **-2**. **Maximum value:** \( 2 \) **Minimum value:** \( -3 \) --- ### (c) What is the average value of \( f \) over \( [0,6] \)? The average value of \( f \) over \( [0,6] \) is: \[ \text{Average} = \frac{1}{6-0} \int_0^6 f(x)dx = \frac{1}{6} \left(y(6) - y(0)\right) \] From the graph: - \( y(0) = 0 \) - \( y(1) = 2 \) - \( y(2) = -1 \) - \( y(4) = 1 \) - \( y(6) = -3 \) So, \[ \int_0^6 f(x)dx = y(6) - y(0) = -3 - 0 = -3 \] So, \[ \text{Average} = \frac{-3}{6} = -\frac{1}{2} \] --- ## **Final Answers:** ### (a) - **\( f \) is positive on:** \( (0,1) \cup (2,4) \) - **\( f \) is negative on:** \( (1,2) \cup (4,6) \) - **\( f = 0 \) at:** \( \{1, 2, 4\} \) ### (b) - **Maximum value:** \( 2 \) - **Minimum value:** \( -3 \) ### (c) - **Average value over \( [0,6] \):** \( -\frac{1}{2} \)