Critical points of f(x, y), classified with the second derivative test. 70 problems with worked solutions; in 70 of them every
equation is proved by a computer algebra system.
20 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - 2 x - y^{2} + 4 y - 3 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - 2 x - y^{2} + 4 y - 3\right)\\\frac{\partial}{\partial y} \left(x^{2} - 2 x - y^{2} + 4 y - 3\right)\end{matrix}\right] = \left[\begin{matrix}2 x - 2\\4 - 2 y\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (1, 2), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (1, 2) \)
A contour map of f(x, y) = x² - 2x - y² + 4y - 3: each curve joins points where f has the same value. The contours close in around the critical point (1, 2).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = - x^{2} - 2 x - 2 y^{2} - 4 y - 5 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(- x^{2} - 2 x - 2 y^{2} - 4 y - 5\right)\\\frac{\partial}{\partial y} \left(- x^{2} - 2 x - 2 y^{2} - 4 y - 5\right)\end{matrix}\right] = \left[\begin{matrix}- 2 x - 2\\- 4 y - 4\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (-1, -1), the only solution.Reviewed
\[ 8 \]
D = f_xx f_yy − f_xy².Reviewed
D = 8 > 0 with f_xx < 0: a local maximum.
Reviewed
Answer \( \text{local maximum at } (-1, -1) \)
A contour map of f(x, y) = -x² - 2x - 2y² - 4y - 5: each curve joins points where f has the same value. The contours close in around the critical point (-1, -1).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = - x^{2} + 4 x - 2 y^{2} + 12 y - 21 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(- x^{2} + 4 x - 2 y^{2} + 12 y - 21\right)\\\frac{\partial}{\partial y} \left(- x^{2} + 4 x - 2 y^{2} + 12 y - 21\right)\end{matrix}\right] = \left[\begin{matrix}4 - 2 x\\12 - 4 y\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (2, 3), the only solution.Reviewed
\[ 8 \]
D = f_xx f_yy − f_xy².Reviewed
D = 8 > 0 with f_xx < 0: a local maximum.
Reviewed
Answer \( \text{local maximum at } (2, 3) \)
A contour map of f(x, y) = -x² + 4x - 2y² + 12y - 21: each curve joins points where f has the same value. The contours close in around the critical point (2, 3).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - 4 x + y^{2} - 2 y + 4 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - 4 x + y^{2} - 2 y + 4\right)\\\frac{\partial}{\partial y} \left(x^{2} - 4 x + y^{2} - 2 y + 4\right)\end{matrix}\right] = \left[\begin{matrix}2 x - 4\\2 y - 2\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (2, 1), the only solution.Reviewed
\[ 4 \]
D = f_xx f_yy − f_xy².Reviewed
D = 4 > 0 with f_xx > 0: a local minimum.
Reviewed
Answer \( \text{local minimum at } (2, 1) \)
A contour map of f(x, y) = x² - 4x + y² - 2y + 4: each curve joins points where f has the same value. The contours close in around the critical point (2, 1).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} + 6 x - y^{2} + 9 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 6 x - y^{2} + 9\right)\\\frac{\partial}{\partial y} \left(x^{2} + 6 x - y^{2} + 9\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 6\\- 2 y\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (-3, 0), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (-3, 0) \)
A contour map of f(x, y) = x² + 6x - y² + 9: each curve joins points where f has the same value. The contours close in around the critical point (-3, 0).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} + 4 x - y^{2} + 6 y - 5 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 4 x - y^{2} + 6 y - 5\right)\\\frac{\partial}{\partial y} \left(x^{2} + 4 x - y^{2} + 6 y - 5\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 4\\6 - 2 y\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (-2, 3), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (-2, 3) \)
A contour map of f(x, y) = x² + 4x - y² + 6y - 5: each curve joins points where f has the same value. The contours close in around the critical point (-2, 3).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} + 6 x + y^{2} + 2 y + 10 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 6 x + y^{2} + 2 y + 10\right)\\\frac{\partial}{\partial y} \left(x^{2} + 6 x + y^{2} + 2 y + 10\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 6\\2 y + 2\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (-3, -1), the only solution.Reviewed
\[ 4 \]
D = f_xx f_yy − f_xy².Reviewed
D = 4 > 0 with f_xx > 0: a local minimum.
Reviewed
Answer \( \text{local minimum at } (-3, -1) \)
A contour map of f(x, y) = x² + 6x + y² + 2y + 10: each curve joins points where f has the same value. The contours close in around the critical point (-3, -1).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
A contour map of f(x, y) = -x² - 2y² + 12y - 20: each curve joins points where f has the same value. The contours close in around the critical point (0, 3).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - y^{2} - 6 y - 9 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - y^{2} - 6 y - 9\right)\\\frac{\partial}{\partial y} \left(x^{2} - y^{2} - 6 y - 9\right)\end{matrix}\right] = \left[\begin{matrix}2 x\\- 2 y - 6\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (0, -3), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (0, -3) \)
A contour map of f(x, y) = x² - y² - 6y - 9: each curve joins points where f has the same value. The contours close in around the critical point (0, -3).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = - x^{2} - 4 x - 2 y^{2} + 4 y - 5 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(- x^{2} - 4 x - 2 y^{2} + 4 y - 5\right)\\\frac{\partial}{\partial y} \left(- x^{2} - 4 x - 2 y^{2} + 4 y - 5\right)\end{matrix}\right] = \left[\begin{matrix}- 2 x - 4\\4 - 4 y\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (-2, 1), the only solution.Reviewed
\[ 8 \]
D = f_xx f_yy − f_xy².Reviewed
D = 8 > 0 with f_xx < 0: a local maximum.
Reviewed
Answer \( \text{local maximum at } (-2, 1) \)
A contour map of f(x, y) = -x² - 4x - 2y² + 4y - 5: each curve joins points where f has the same value. The contours close in around the critical point (-2, 1).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - 6 x + y^{2} + 4 y + 15 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - 6 x + y^{2} + 4 y + 15\right)\\\frac{\partial}{\partial y} \left(x^{2} - 6 x + y^{2} + 4 y + 15\right)\end{matrix}\right] = \left[\begin{matrix}2 x - 6\\2 y + 4\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (3, -2), the only solution.Reviewed
\[ 4 \]
D = f_xx f_yy − f_xy².Reviewed
D = 4 > 0 with f_xx > 0: a local minimum.
Reviewed
Answer \( \text{local minimum at } (3, -2) \)
A contour map of f(x, y) = x² - 6x + y² + 4y + 15: each curve joins points where f has the same value. The contours close in around the critical point (3, -2).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} + 6 x + y^{2} - 6 y + 15 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 6 x + y^{2} - 6 y + 15\right)\\\frac{\partial}{\partial y} \left(x^{2} + 6 x + y^{2} - 6 y + 15\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 6\\2 y - 6\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (-3, 3), the only solution.Reviewed
\[ 4 \]
D = f_xx f_yy − f_xy².Reviewed
D = 4 > 0 with f_xx > 0: a local minimum.
Reviewed
Answer \( \text{local minimum at } (-3, 3) \)
A contour map of f(x, y) = x² + 6x + y² - 6y + 15: each curve joins points where f has the same value. The contours close in around the critical point (-3, 3).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Both vanish at (0, -2), the only solution.Reviewed
\[ 8 \]
D = f_xx f_yy − f_xy².Reviewed
D = 8 > 0 with f_xx < 0: a local maximum.
Reviewed
Answer \( \text{local maximum at } (0, -2) \)
A contour map of f(x, y) = -x² - 2y² - 8y - 8: each curve joins points where f has the same value. The contours close in around the critical point (0, -2).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - 6 x - y^{2} + 9 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - 6 x - y^{2} + 9\right)\\\frac{\partial}{\partial y} \left(x^{2} - 6 x - y^{2} + 9\right)\end{matrix}\right] = \left[\begin{matrix}2 x - 6\\- 2 y\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (3, 0), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (3, 0) \)
A contour map of f(x, y) = x² - 6x - y² + 9: each curve joins points where f has the same value. The contours close in around the critical point (3, 0).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Both vanish at (-1, 1), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (-1, 1) \)
A contour map of f(x, y) = x² + 2x - y² + 2y: each curve joins points where f has the same value. The contours close in around the critical point (-1, 1).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} + 2 x - y^{2} - 4 y - 3 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 2 x - y^{2} - 4 y - 3\right)\\\frac{\partial}{\partial y} \left(x^{2} + 2 x - y^{2} - 4 y - 3\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 2\\- 2 y - 4\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (-1, -2), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (-1, -2) \)
A contour map of f(x, y) = x² + 2x - y² - 4y - 3: each curve joins points where f has the same value. The contours close in around the critical point (-1, -2).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} + 6 x - y^{2} + 4 y + 5 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 6 x - y^{2} + 4 y + 5\right)\\\frac{\partial}{\partial y} \left(x^{2} + 6 x - y^{2} + 4 y + 5\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 6\\4 - 2 y\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (-3, 2), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (-3, 2) \)
A contour map of f(x, y) = x² + 6x - y² + 4y + 5: each curve joins points where f has the same value. The contours close in around the critical point (-3, 2).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - 2 x - y^{2} - 6 y - 8 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - 2 x - y^{2} - 6 y - 8\right)\\\frac{\partial}{\partial y} \left(x^{2} - 2 x - y^{2} - 6 y - 8\right)\end{matrix}\right] = \left[\begin{matrix}2 x - 2\\- 2 y - 6\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (1, -3), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (1, -3) \)
A contour map of f(x, y) = x² - 2x - y² - 6y - 8: each curve joins points where f has the same value. The contours close in around the critical point (1, -3).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - 6 x - y^{2} + 2 y + 8 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - 6 x - y^{2} + 2 y + 8\right)\\\frac{\partial}{\partial y} \left(x^{2} - 6 x - y^{2} + 2 y + 8\right)\end{matrix}\right] = \left[\begin{matrix}2 x - 6\\2 - 2 y\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (3, 1), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (3, 1) \)
A contour map of f(x, y) = x² - 6x - y² + 2y + 8: each curve joins points where f has the same value. The contours close in around the critical point (3, 1).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} - 4 x - y^{2} - 2 y + 3 \).
\[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} - 4 x - y^{2} - 2 y + 3\right)\\\frac{\partial}{\partial y} \left(x^{2} - 4 x - y^{2} - 2 y + 3\right)\end{matrix}\right] = \left[\begin{matrix}2 x - 4\\- 2 y - 2\end{matrix}\right] \]
The partial derivatives.✓ Proved
\[ \left[\begin{matrix}0\\0\end{matrix}\right] \]
Both vanish at (2, -1), the only solution.Reviewed
\[ -4 \]
D = f_xx f_yy − f_xy².Reviewed
D = -4 < 0: a saddle point.
Reviewed
Answer \( \text{saddle point at } (2, -1) \)
A contour map of f(x, y) = x² - 4x - y² - 2y + 3: each curve joins points where f has the same value. The contours close in around the critical point (2, -1).
✓ Nihil obstat Lines: 1 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify