1. **Problem Statement:** We are given the graph of $f'$, the derivative of a function $f$, and need to determine the intervals on which $f(x)$ is concave up on the open interval $(-9,9)$. 2. **Recall the relationship between concavity and derivatives:** - The concavity of $f$ depends on the sign of the second derivative $f''(x)$. - If $f''(x) > 0$ on an interval, then $f$ is concave up there. - Since $f''(x) = (f'(x))'$, the second derivative is the derivative of $f'$. 3. **Interpretation using the graph of $f'$:** - $f''(x) > 0$ means $f'(x)$ is increasing. - $f''(x) < 0$ means $f'(x)$ is decreasing. 4. **Analyze the graph of $f'$ given:** - From $x=-9$ to $x=-7$, $f'(x)$ is decreasing (from about 5 to -4), so $f''(x)<0$ (concave down). - From $x=-7$ to $x=-4$, $f'(x)$ is increasing (from -4 to 1), so $f''(x)>0$ (concave up). - From $x=-4$ to $x=-1$, $f'(x)$ is decreasing (from 1 to -5), so $f''(x)<0$ (concave down). - From $x=-1$ to $x=3$, $f'(x)$ is increasing (from -5 to 2), so $f''(x)>0$ (concave up). - From $x=3$ to $x=9$, $f'(x)$ is decreasing (from 2 to -2), so $f''(x)<0$ (concave down). 5. **Conclusion:** - $f$ is concave up on intervals where $f'(x)$ is increasing: $$(-7,-4) \cup (-1,3)$$ **Final answer:** The function $f$ is concave up on the intervals $$(-7,-4) \text{ and } (-1,3).$$