1. **State the problem:** We need to sketch a function $f$ continuous on $[1,5]$ with a second derivative on $(1,5)$ satisfying: - $f(1)=f(5)=3$, $f(2)=0$, $f(3)=1$, $f(4)=2$ - $f'(2)=f'(4)=0$ - $f'(x)<0$ on $(1,2)$, $f'(x)>0$ on $(2,4)$ and $(4,5)$ - $f''(x)>0$ on $(1,2)$ and $(4,5)$, $f''(x)<0$ on $(3,4)$ 2. **Analyze the function values and derivatives:** - At $x=1$ and $x=5$, $f=3$. - The function decreases from $3$ at $x=1$ to $0$ at $x=2$ since $f'(x)<0$ on $(1,2)$. - At $x=2$, $f'(2)=0$ (horizontal tangent), then $f'(x)>0$ on $(2,4)$, so the function increases from $0$ at $x=2$ to $2$ at $x=4$. - At $x=4$, $f'(4)=0$ again, then $f'(x)>0$ on $(4,5)$, so the function continues increasing from $2$ at $x=4$ to $3$ at $x=5$. 3. **Analyze concavity using second derivative:** - $f''(x)>0$ on $(1,2)$ and $(4,5)$ means the graph is concave up (shaped like a cup) on these intervals. - $f''(x)<0$ on $(3,4)$ means the graph is concave down (shaped like a cap) on this interval. - Between $2$ and $3$, the concavity is not specified, but since $f''$ changes sign, it likely transitions. 4. **Sketching the graph:** - From $x=1$ to $x=2$, the function decreases with concave up shape, going from $3$ to $0$. - At $x=2$, the slope is zero (horizontal tangent). - From $x=2$ to $x=3$, the function increases, concavity likely changes from up to down. - From $x=3$ to $x=4$, the function increases with concave down shape, going from $1$ to $2$. - At $x=4$, slope zero again. - From $x=4$ to $x=5$, the function increases with concave up shape, going from $2$ to $3$. 5. **Summary:** The function starts at $(1,3)$, dips down to $(2,0)$ with a concave up shape, then rises through $(3,1)$ and $(4,2)$ with a concave down shape between $3$ and $4$, and finally rises to $(5,3)$ with concave up shape. This satisfies all given conditions. **Final answer:** The graph is a continuous curve on $[1,5]$ with horizontal tangents at $x=2$ and $x=4$, decreasing then increasing slopes, and changing concavity as specified.