1. **Problem statement:** We have a function $f$ defined graphically on $[-10,10]$ with line segments and a quarter circle. The function $g$ is defined as $$g(x) = \int_1^x f(t) \, dt.$$ We need to find all intervals on $[-9,9]$ where $g$ is concave down. 2. **Recall the relationship between $g$ and $f$:** Since $g(x) = \int_1^x f(t) \, dt$, by the Fundamental Theorem of Calculus, $$g'(x) = f(x).$$ 3. **Concavity and second derivative:** The concavity of $g$ depends on the sign of $g''(x)$: - If $g''(x) < 0$, then $g$ is concave down. - If $g''(x) > 0$, then $g$ is concave up. Since $g'(x) = f(x)$, then $$g''(x) = f'(x).$$ 4. **Interpretation:** To find where $g$ is concave down, we need to find where $f'(x) < 0$ on $[-9,9]$. 5. **Analyze $f$ on $[-9,9]$ using the graph segments:** - From $-9$ to $-6$: $f$ is on the line segment from $(-10,0)$ to $(-6,6)$, which is increasing (slope positive). - From $-6$ to $-1$: $f$ goes from $( -6,6)$ to $(-1,0)$, which is decreasing (slope negative). - From $-1$ to $1$: $f$ is constant at $0$ (slope zero). - From $1$ to $3$: $f$ goes from $(1,0)$ to $(3,4)$, increasing (slope positive). - From $3$ to $5$: $f$ goes from $(3,4)$ to $(5,-5)$, decreasing (slope negative). - From $5$ to $7$: $f$ goes from $(5,-5)$ to $(7,3)$, increasing (slope positive). - From $7$ to $9$: $f$ follows a quarter circle arc from $(7,3)$ to $(9,5)$, which is concave down but we need the slope of $f$ (derivative of $f$). The quarter circle is the top-right quarter of a circle rising, so the slope $f'(x)$ is decreasing (negative) because the circle's slope decreases as $x$ increases in this arc. 6. **Summarize intervals where $f'(x) < 0$ (i.e., $g$ concave down):** - $(-6,-1)$: $f$ decreasing - $(3,5)$: $f$ decreasing - $(7,9)$: slope of $f$ decreasing on quarter circle arc 7. **Final answer:** $$\boxed{\text{The function } g \text{ is concave down on } [-9,9] \text{ precisely on the intervals } (-6,-1), (3,5), \text{ and } (7,9).}$$