1. **Problem Statement:** Determine the signs of the partial derivatives $f_x(2,-2)$, $f_{xx}(2,-2)$, $f_y(2,-2)$, and $f_{yy}(2,-2)$ for the given saddle-shaped surface symmetric about the origin. 2. **Understanding the graph:** The surface has peaks near $(-2,2)$ and $(2,-2)$ and valleys near $(2,2)$ and $(-2,-2)$. The point $(2,-2)$ is near a positive peak. 3. **Partial derivative $f_x(2,-2)$:** This derivative measures the slope of the surface in the $x$-direction at $(2,-2)$. - Since $(2,-2)$ is near a peak, moving in the $x$-direction around this point will show the slope going from lower to higher values or vice versa. - At a peak, the slope changes from positive to negative, so the first derivative $f_x$ at the peak is zero or close to zero. - But since the peak is near $(2,-2)$, and the function is symmetric, the slope $f_x(2,-2)$ is approximately zero. 4. **Second partial derivative $f_{xx}(2,-2)$:** This measures the concavity in the $x$-direction. - At a peak, the surface is concave down, so $f_{xx}(2,-2) < 0$ (negative). 5. **Partial derivative $f_y(2,-2)$:** This measures the slope in the $y$-direction at $(2,-2)$. - Since $(2,-2)$ is a peak, the slope in $y$-direction is zero or close to zero. 6. **Second partial derivative $f_{yy}(2,-2)$:** Measures concavity in $y$-direction. - At a peak, concavity is downwards, so $f_{yy}(2,-2) < 0$ (negative). 7. **Summary:** - $f_x(2,-2) \approx 0$ - $f_{xx}(2,-2) < 0$ - $f_y(2,-2) \approx 0$ - $f_{yy}(2,-2) < 0$ 8. **Matching to options:** The closest match is (F) zero, zero for first derivatives and (E) negative, negative for second derivatives. Since the problem asks for pairs, the answers are: - (a) $f_x(2,-2)$ and $f_{xx}(2,-2)$: zero, negative (closest to (F) zero, zero or (E) negative, negative but zero, negative fits best) - (b) $f_y(2,-2)$ and $f_{yy}(2,-2)$: zero, negative (matches (I) zero, negative) **Final answer:** (a) $f_x(2,-2) = 0$, $f_{xx}(2,-2) < 0$ (b) $f_y(2,-2) = 0$, $f_{yy}(2,-2) < 0$