1. The problem asks to find all intervals where the function $f(x)$ is concave down on the open interval $(-9,9)$. 2. Recall that concavity is determined by the second derivative $f''(x)$. If $f''(x)<0$ on an interval, then $f$ is concave down there. 3. The graph shows $f(x)$ and the derivative $f'(x)$ plotted in real time. Since $f'(x)$ is the slope of $f$, the concavity depends on whether $f'(x)$ is increasing or decreasing. 4. Specifically, $f''(x)$ is the derivative of $f'(x)$. So if $f'(x)$ is decreasing, then $f''(x)<0$ and $f$ is concave down. 5. From the description, $f'(x)$ at $x=-8.9$ is negative (-1.00124), and the graph of $f$ rises from $x=-9$ to about $x=-4$, then dips near $x=0$, then rises sharply to $x=6$, then descends near $x=9$. 6. Observing the slope behavior: - From $x=-9$ to about $x=-4$, the slope $f'(x)$ increases (since $f$ is rising less steeply), so $f''(x)>0$ (concave up). - From about $x=-4$ to $x=0$, the slope $f'(x)$ decreases (since $f$ dips), so $f''(x)<0$ (concave down). - From $x=0$ to $x=6$, the slope $f'(x)$ increases sharply (rising steeply), so $f''(x)>0$ (concave up). - From $x=6$ to $x=9$, the slope $f'(x)$ decreases (descending slightly), so $f''(x)<0$ (concave down). 7. Therefore, the intervals where $f$ is concave down are approximately $(-4,0)$ and $(6,9)$. **Final answer:** $$\boxed{(-4,0) \text{ and } (6,9)}$$