Extrema & Saddle Points - Calculus 3

1. A function $h(x,y)$ has critical points at $A(1,2)$, $B(-1,0)$, and $C(0,1)$. The second derivative test gives: at $A$: $h_{xx} = 4$, $h_{yy} = 6$, $h_{xy} = -1$; at $B$: $h_{xx} = -2$, $h_{yy} = -3$, $h_{xy} = 2$; at $C$: $h_{xx} = 2$, $h_{yy} = -1$, $h_{xy} = 3$. How many local extrema does the function have?
   Answer: Two local extrema: one local minimum and one local maximum Apply the second derivative test at each point using $D = h_{xx}h_{yy} - (h_{xy})^2$. At $A(1,2)$: $D = (4)(6) - (-1)^2 = 24 - 1 = 23 > 0$ and $h_{xx} = 4 > 0$, so local minimum. At $B(-1,0)$: $D = (-2)(-3) - (2)^2 = 6 - 4 = 2 > 0$ and $h_{xx} = -2 < 0$, so local maximum. At $C(0,1)$: $D = (2)(-1) - (3)^2 = -2 - 9 = -11 < 0$, so saddle point. Therefore, there are exactly two local extrema.
2. The function $F(x,y) = e^{x^2-y^2}\sin(xy)$ has a critical point at the origin. To determine its nature, which approach would be most effective given that the second derivative test yields $D = 0$?
   Answer: Analyze the sign of $F(x,y)$ along the coordinate axes and diagonal lines $y = ±x$ When $D = 0$ at a critical point, examining the function's behavior along specific directions is often the most practical approach. For $F(x,y) = e^{x^2-y^2}\sin(xy)$, note that $F(0,0) = e^0\sin(0) = 0$. Along the $x$-axis ($y = 0$): $F(x,0) = e^{x^2}\sin(0) = 0$. Along the $y$-axis ($x = 0$): $F(0,y) = e^{-y^2}\sin(0) = 0$. Along $y = x$: $F(x,x) = e^{x^2-x^2}\sin(x^2) = \sin(x^2)$, which changes sign near the origin. Along $y = -x$: $F(x,-x) = e^{x^2-x^2}\sin(-x^2) = -\sin(x^2)$. Since the function takes both positive and negative values in any neighborhood of the origin, it's a saddle point.
3. Consider the function $H(x,y) = (x^2 + y^2 - 1)^2 + (x - y)^2$. When analyzing this function for critical points, what is the most significant challenge in applying the second derivative test?
   Answer: Multiple critical points yield $D = 0$, requiring alternative classification methods To find critical points, we solve $H_x = 4(x^2 + y^2 - 1)x + 2(x - y) = 0$ and $H_y = 4(x^2 + y^2 - 1)y - 2(x - y) = 0$. The critical points include $(0,0)$ and points where $x = y$ and $x^2 + y^2 = 1$, namely $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$. The second derivatives are $H_{xx} = 4(x^2 + y^2 - 1) + 8x^2 + 2$, $H_{yy} = 4(x^2 + y^2 - 1) + 8y^2 + 2$, and $H_{xy} = 8xy - 2$. At the points where $x^2 + y^2 = 1$, the discriminant $D = H_{xx}H_{yy} - (H_{xy})^2$ often equals zero, making the standard second derivative test inconclusive.
4. Consider $f(x,y) = ax^2 + bxy + cy^2 + dx + ey + k$ where $a, b, c$ are constants with $a \neq 0$ and $c \neq 0$. If this function has exactly one critical point and it is a saddle point, which condition must be satisfied?
   Answer: $4ac - b^2 < 0$ and the system $\nabla f = 0$ has a unique solution When analyzing critical points of multivariable functions, you need to find where the gradient equals zero, then use the second derivative test to classify each critical point. For quadratic functions like this one, the analysis becomes systematic. To find critical points, set $\nabla f = 0$: $f_x = 2ax + by + d = 0$ and $f_y = bx + 2cy + e = 0$. This gives you a system of two linear equations. Since $a \neq 0$ and $c \neq 0$, this system has exactly one solution when the coefficient matrix is non-singular, which it is here. The second derivative test uses the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$. For your function: $f_{xx} = 2a$, $f_{yy} = 2c$, and $f_{xy} = b$. Therefore: $D = (2a)(2c) - b^2 = 4ac - b^2$. For a saddle point, you need $D < 0$, which means $4ac - b^2 < 0$. The system $\nabla f = 0$ indeed has a unique solution since both $a$ and $c$ are non-zero. This confirms answer C. Answer A describes the condition for a local extremum (minimum or maximum), not a saddle point. Answer B suggests the test is inconclusive, but $D = 0$ would require additional analysis beyond the standard second derivative test. Answer D is incomplete because it ignores the discriminant relationship—you need the full condition $4ac - b^2 < 0$, not just $ac < 0$. Study tip: Remember the second derivative test discriminant: $D < 0$ means saddle point, $D > 0$ means local extremum.
5. A function $K(x,y) = x^4 + y^4 + ax^2y^2 + bx^2 + cy^2$ has the origin as its only critical point. If this critical point is classified as a local minimum, what constraint must the parameter $a$ satisfy?
   Answer: $a > -2$ to ensure the function is positive definite near the origin For the origin to be the only critical point, we need $\nabla K = (4x^3 + 2ax y^2 + 2bx, 4y^3 + 2axy^2 + 2cy) = (0,0)$ only at $(0,0)$. This requires $b = c = 0$, giving us $K(x,y) = x^4 + y^4 + ax^2y^2$. For this to have a local minimum at the origin, we need $K(x,y) \geq 0$ in a neighborhood of $(0,0)$. We can write this as a quadratic form in $u = x^2$ and $v = y^2$: $K = u^2 + v^2 + auv$ where $u,v \geq 0$. This quadratic form is positive semidefinite when its discriminant satisfies $4 - a^2 \geq 0$, which gives us $a^2 \leq 4$, or equivalently $a \geq -2$. Since we need strict inequality for a minimum, $a > -2$.
6. For the function $L(x,y) = xy\ln(x^2 + y^2)$ defined for $(x,y) \neq (0,0)$, consider the behavior near the origin. If we were to extend this function continuously to include the origin, what type of critical point would $(0,0)$ represent?
   Answer: A saddle point since $L$ takes both positive and negative values arbitrarily close to the origin When analyzing critical points of multivariable functions, especially near discontinuities or points where the function isn't originally defined, you need to examine how the function behaves in different directions approaching that point. To classify what type of critical point (0,0) would be if we extend $L(x,y) = xy\ln(x^2 + y^2)$ continuously, we must determine whether the function takes both positive and negative values near the origin. The sign of $L(x,y)$ depends on two factors: the sign of $xy$ and the sign of $\ln(x^2 + y^2)$. Since $x^2 + y^2 < 1$ when we're close to the origin, we have $\ln(x^2 + y^2) < 0$. The term $xy$ is positive in the first and third quadrants, negative in the second and fourth quadrants. Therefore, $L(x,y)$ is negative in quadrants I and III, and positive in quadrants II and IV, arbitrarily close to the origin. Answer A is incorrect because while the limit is indeed 0, the fact that $L$ changes sign actually indicates a saddle point, not a minimum. Answer B is wrong because there's no maximum behavior—the function takes both positive and negative values nearby. Answer D misses the point entirely; computing the limit (which is 0) doesn't resolve the classification—we need the sign analysis. The correct answer is C: since $L$ takes both positive and negative values in every neighborhood of the origin, (0,0) represents a saddle point. Study tip: For critical point classification near discontinuities, always check the function's sign in different directions—if it's both positive and negative nearby, you've found a saddle point.
7. Consider the function $f(x,y) = x^3 + y^3 - 3xy + 2$. After finding all critical points and applying the second derivative test, what can be concluded about the nature of the critical points?
   Answer: There are two critical points: $(0,0)$ is a saddle point and $(1,1)$ is a local minimum First, find critical points by setting $f_x = 3x^2 - 3y = 0$ and $f_y = 3y^2 - 3x = 0$. From the first equation, $y = x^2$. Substituting into the second: $3(x^2)^2 - 3x = 3x^4 - 3x = 3x(x^3 - 1) = 0$. This gives $x = 0$ or $x = 1$, so critical points are $(0,0)$ and $(1,1)$. For the second derivative test: $f_{xx} = 6x$, $f_{yy} = 6y$, $f_{xy} = -3$. At $(0,0)$: $D = (0)(0) - (-3)^2 = -9 < 0$, so it's a saddle point. At $(1,1)$: $D = (6)(6) - (-3)^2 = 27 > 0$ and $f_{xx} = 6 > 0$, so it's a local minimum.

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