1. **Problem Statement:** We are given the graph of the derivative $f'$ of a function $f$ on the interval $[-9,9]$ and the value $f(9) = -2$. We need to find the maximum value of $f$ on the closed interval $[-9,9]$. 2. **Key Idea:** The function $f$ increases where $f' > 0$ and decreases where $f' < 0$. To find the maximum of $f$, we look for critical points where $f' = 0$ or where $f'$ changes sign, and also check the endpoints. 3. **Given Points of $f'$:** - $f'(-9) = -2$ (negative) - $f'(-5) = 4$ (positive) - $f'(1) = 4$ (positive) - $f'(6) = -6$ (negative) - $f'(9) = 6$ (positive) 4. **Intervals of Increase/Decrease:** - From $-9$ to $-5$, $f'$ goes from negative to positive, so $f$ is decreasing then increasing, indicating a local minimum near $x=-9$ or $x=-5$. - From $-5$ to $1$, $f' > 0$, so $f$ is increasing. - From $1$ to $6$, $f'$ changes from positive to negative, so $f$ increases then decreases, indicating a local maximum near $x=1$ or $x=6$. - From $6$ to $9$, $f'$ changes from negative to positive, so $f$ decreases then increases, indicating a local minimum near $x=6$ or $x=9$. 5. **Calculate $f$ values using the Fundamental Theorem of Calculus:** We know $f(9) = -2$. We can find $f$ at other points by integrating $f'$ backward from $9$ to those points: $$f(a) = f(9) - \int_a^9 f'(x) \, dx$$ 6. **Approximate the integral using trapezoids between given points:** - From $9$ to $6$: $$\int_6^9 f'(x) dx \approx \frac{(f'(6) + f'(9))}{2} \times (9-6) = \frac{-6 + 6}{2} \times 3 = 0$$ - From $6$ to $1$: $$\int_1^6 f'(x) dx \approx \frac{(f'(1) + f'(6))}{2} \times (6-1) = \frac{4 + (-6)}{2} \times 5 = \frac{-2}{2} \times 5 = -5$$ - From $1$ to $-5$: $$\int_{-5}^1 f'(x) dx \approx \frac{(f'(-5) + f'(1))}{2} \times (1 - (-5)) = \frac{4 + 4}{2} \times 6 = 4 \times 6 = 24$$ - From $-5$ to $-9$: $$\int_{-9}^{-5} f'(x) dx \approx \frac{(f'(-9) + f'(-5))}{2} \times (-5 - (-9)) = \frac{-2 + 4}{2} \times 4 = 1 \times 4 = 4$$ 7. **Calculate $f$ at key points:** - $f(6) = f(9) - \int_6^9 f'(x) dx = -2 - 0 = -2$ - $f(1) = f(6) - \int_1^6 f'(x) dx = -2 - (-5) = -2 + 5 = 3$ - $f(-5) = f(1) - \int_{-5}^1 f'(x) dx = 3 - 24 = -21$ - $f(-9) = f(-5) - \int_{-9}^{-5} f'(x) dx = -21 - 4 = -25$ 8. **Determine the maximum value:** From the values: - $f(-9) = -25$ - $f(-5) = -21$ - $f(1) = 3$ - $f(6) = -2$ - $f(9) = -2$ The maximum value of $f$ on $[-9,9]$ is $3$ at $x=1$. **Final answer:** $$\boxed{3}$$