1. The problem asks to find the value of $x$ where the function $h(x)$ is continuous but not differentiable. 2. A function is continuous at $x=a$ if the left-hand limit, right-hand limit, and the function value at $a$ are all equal. 3. A function is not differentiable at $x=a$ if there is a sharp corner or cusp, meaning the left-hand and right-hand derivatives at $a$ are not equal. 4. From the graph points: - At $x=2$, $h(2)=1$ (filled point), but there is an open circle at $(2,2)$, so the function jumps from 1 to 2, not continuous. - At $x=3$, $h(3)=2$ (filled point), and there is an open circle at $(3,1)$, so the function jumps from 2 to 1, not continuous. - At $x=4$, $h(4)=1$ (filled point), no open circle at 4, so continuous. - At $x=5$, there is an open circle at $(5,0.5)$ but no filled point, so function is not defined there, not continuous. 5. Check differentiability at $x=4$: - The slope from $x=3$ to $x=4$ is $\frac{1-2}{4-3} = -1$. - The slope from $x=4$ to $x=6$ is $\frac{0-1}{6-4} = -\frac{1}{2}$. 6. Since the left and right slopes at $x=4$ are different ($-1 \neq -\frac{1}{2}$), $h(x)$ is not differentiable at $x=4$. 7. Therefore, $h(x)$ is continuous but not differentiable at $x=4$. **Final answer:** $\boxed{4}$