1. **State the problems:** We have two graphs with their skeletons of function $f(x)$, and we want to match the derivative $f'(x)$ for each based only on the sign of $f'(x)$ in the interval $(-1.5, 1.5)$. --- 2. **Analyze Graph 1 and its derivative $f'(x) = 1 - x^2$: ** - $f'(x) = 1 - x^2$ is a downward parabola with zeros at $x = \pm 1$. - On $(-1.5, -1)$, $f'(x) < 0$ since $x^2 > 1$. - On $(-1, 1)$, $f'(x) > 0$ since $x^2 < 1$. - On $(1, 1.5)$, $f'(x) < 0$ again. - At $x = -1$ and $x=1$, $f'(x) = 0$. This matches the description: decreasing to zero at $x=-1$, increasing to zero at $x=1$, then decreasing again. --- 3. **Analyze Graph 2 and its derivative $f'(x) = (1 + x)^2$: ** - Since $(1+x)^2 \geq 0$ for all $x$, $f'(x)$ is never negative. - On $(-1.5, -1)$: $f'(x) > 0$ as $(1+x)^2$ is positive. - At $x=-1$, $f'(-1) = 0$. - On $(-1, 1.5)$, $f'(x) > 0$. This corresponds to flat at $y=0$ on $[-1, 1]$ (since derivative is zero at $x=-1$ and positive just after) and increasing after $x=1$. --- 4. **Final answer:** - Graph 1 corresponds to $f'(x) = 1 - x^2$. - Graph 2 corresponds to $f'(x) = (1 + x)^2$. These matches are consistent with the signs and zeros of the derivatives and the shape given.