Let's break down the problem into two steps as requested. --- ## **Step 1: Find \( f() \) and \( f(7) \) Using the Graph of \( f'(x) \)** ### Explanation: - The graph shows \('(x) \) is **piecewise constant**: - From \( x = \) to \( x = 2 \), \( f'(x) = -3 \) - From \( x = 2 \) to \( x = 5 \), \( f'(x) = 1 \) - From \( x = 5 \) to \( x = 7 \), \( f'(x) = 1 \) - We know **\( f(5) = \)**. - To find \( f() \) and \( f(7) \), use the Fundamental Theorem of Calculus: \[ f(b) = f(a) + \int_a^b f'(x) \, dx \] #### **Finding \( f() \):** We know \( f(5) = \), want to find \( f() \): \[ f(5) = f() + \int_^5 f'(x) \, dx \] So, \[ f() = f(5) - \int_^5 f'(x) \, dx \] Calculate the integral: - From to 2: \( f'(x) = -3 \) ⇒ Area = \( -3 \times 2 = -6 \) - From 2 to 5: \( f'(x) = 1 \) ⇒ Area = \( 1 \times 3 = 3 \) - Total: \( -6 + 3 = -3 \) So: \[ f() = - (-3) = 3 \] **Explanation:** We found \( f() \) by integrating \( f'(x) \) from to 5 and subtracting from \( f(5) \). --- #### **Finding \( f(7) \):** Similarly, \[ f(7) = f(5) + \int_5^7 f'(x) \, dx \] From 5 to 7: \( f'(x) = 1 \) ⇒ Area = \( 1 \times 2 = 2 \) So: \[ f(7) = + 2 = 2 \] **Explanation:** We found \( f(7) \) by integrating \( f'(x) \) from 5 to 7 and adding to \( f(5) \). --- ## **Step 2: Calculate \( \int_^7 f'(x) \, dx \)** ### Explanation: The total integral is the sum of areas under \( f'(x) \) from to 7: - From to 2: \( -3 \times 2 = -6 \) - From 2 to 5: \( 1 \times 3 = 3 \) - From 5 to 7: \( 1 \times 2 = 2 \) Sum: \( -6 + 3 + 2 = -1 \) Alternatively, by the Fundamental Theorem: \[ \int_^7 f'(x) dx = f(7) - f() = 2 - 3 = -1 \] **Explanation:** The value of the integral is the net area under the graph between \( x = \) and \( x = 7 \), or simply \( f(7) - f() \). --- ## **Final Answers** - \( f() = \boxed{3} \) - \( f(7) = \boxed{2} \) - \( \int_^7 f'(x) dx = \boxed{-1} \)