Finding Critical Points - Multivariable Calculus

1. How many critical points does the function $f(x, y) = x e^{x^2 - 3y^2}$ have?
   Answer: 0 To find the critical points, we set the first partial derivatives of $f(x, y)$ to zero. $f_x(x, y) = e^{x^2 - 3y^2} + x e^{x^2 - 3y^2}(2x) = (1 + 2x^2)e^{x^2 - 3y^2}$. $f_y(x, y) = x e^{x^2 - 3y^2}(-6y) = -6xy e^{x^2 - 3y^2}$. For a critical point, both must be zero. Since $e^{x^2 - 3y^2}$ is always positive, we can divide it out. The equations become: 1) $1 + 2x^2 = 0$ 2) $-6xy = 0$ From equation (1), $2x^2 = -1$, which has no real solutions for $x$. Therefore, there is no real pair $(x, y)$ that can satisfy the first equation, and the function has no critical points.
2. Find the number of critical points of the function $f(x, y) = \frac{xy}{x^2 + y^2 + 1}$.
   Answer: 1 When finding critical points of multivariable functions, you need to locate where both partial derivatives equal zero simultaneously. Critical points occur where the gradient vector $\nabla f = \langle f_x, f_y \rangle = \langle 0, 0 \rangle$. For $f(x, y) = \frac{xy}{x^2 + y^2 + 1}$, let's compute the partial derivatives using the quotient rule: $f_x = \frac{y(x^2 + y^2 + 1) - xy(2x)}{(x^2 + y^2 + 1)^2} = \frac{y(y^2 + 1 - x^2)}{(x^2 + y^2 + 1)^2}$ $f_y = \frac{x(x^2 + y^2 + 1) - xy(2y)}{(x^2 + y^2 + 1)^2} = \frac{x(x^2 + 1 - y^2)}{(x^2 + y^2 + 1)^2}$ Setting both equal to zero: since the denominators are always positive, we need the numerators to be zero. From $f_x = 0$: either $y = 0$ or $y^2 + 1 - x^2 = 0$ From $f_y = 0$: either $x = 0$ or $x^2 + 1 - y^2 = 0$ If $y = 0$ and $x = 0$, we get the point $(0,0)$. If $y^2 + 1 - x^2 = 0$ and $x^2 + 1 - y^2 = 0$, adding these equations gives $2 = 0$, which is impossible. Therefore, the only critical point is $(0,0)$. A) 5 and B) 3 suggest multiple critical points, likely from incorrectly solving the system of equations or missing the contradiction when trying to solve $y^2 + 1 = x^2$ and $x^2 + 1 = y^2$ simultaneously. D) Infinitely many might result from thinking the equations have parametric solutions, but the algebraic contradiction eliminates this possibility. Strategy tip: Always check that your candidate critical points actually satisfy both partial derivative equations simultaneously—many apparent solutions create contradictions when substituted back.
3. For the function $g(x,y) = \frac{x^2 + y^2}{x^2 + y^2 + 1}$, which statement about critical points is true?
   Answer: The function has exactly one critical point at the origin Let $u = x^2 + y^2$. Then $g(x,y) = \frac{u}{u+1}$. Computing partial derivatives: $g_x = \frac{\partial}{\partial x}\left[\frac{x^2+y^2}{x^2+y^2+1}\right]$. Using the quotient rule: $g_x = \frac{2x(x^2+y^2+1) - (x^2+y^2)(2x)}{(x^2+y^2+1)^2} = \frac{2x(x^2+y^2+1-x^2-y^2)}{(x^2+y^2+1)^2} = \frac{2x}{(x^2+y^2+1)^2}$. Similarly, $g_y = \frac{2y}{(x^2+y^2+1)^2}$. For critical points, we need both partial derivatives to equal zero: $\frac{2x}{(x^2+y^2+1)^2} = 0$ and $\frac{2y}{(x^2+y^2+1)^2} = 0$. Since the denominator $(x^2+y^2+1)^2 > 0$ always, we need $x = 0$ and $y = 0$. Therefore, the only critical point is at the origin $(0,0)$.
4. A surface is defined by the equation $x^2 + 2y^2 + z^2 - 2xy - 2z = 3$. At which $(x,y)$ point(s) is the tangent plane to this surface horizontal?
   Answer: At $(0, 0)$ only. When you encounter a question about horizontal tangent planes to surfaces, you're working with the fundamental concept that a tangent plane is horizontal when its normal vector points straight up or down (parallel to the z-axis). For a surface defined by $F(x,y,z) = x^2 + 2y^2 + z^2 - 2xy - 2z - 3 = 0$, the normal vector is the gradient: $\nabla F = (2x - 2y, 4y - 2x, 2z - 2)$. A horizontal tangent plane occurs when the x and y components of this normal vector are zero, meaning: - $2x - 2y = 0 \Rightarrow x = y$ - $4y - 2x = 0 \Rightarrow 2y = x$ Solving this system: from the first equation $x = y$, substituting into the second gives $2y = y$, so $y = 0$, which means $x = 0$. Therefore, the only point where both conditions are satisfied is $(0,0)$. Choice A is incorrect because while $(0,0)$ works, $(1,1)$ doesn't satisfy the second condition: $4(1) - 2(1) = 2 \neq 0$. Choice B is wrong because at $(2,1)$: $2(2) - 2(1) = 2 \neq 0$ and $4(1) - 2(2) = 0$, so only one condition is met. Similarly for $(-2,-1)$. Choice D incorrectly suggests an entire ellipse of solutions, but our system of equations yields only one point. Study tip: For horizontal tangent planes, always set the partial derivatives with respect to x and y equal to zero simultaneously. Don't forget that both conditions must be satisfied at the same point.
5. How many critical points does the function $f(x, y) = x^4 + y^3 - 4x^2 - 3y^2$ have?
   Answer: 6 When finding critical points of multivariable functions, you need to locate where both partial derivatives equal zero simultaneously. Critical points occur where the gradient vector $\nabla f = \langle f_x, f_y \rangle = \langle 0, 0 \rangle$. For $f(x, y) = x^4 + y^3 - 4x^2 - 3y^2$, let's find the partial derivatives: - $f_x = 4x^3 - 8x = 4x(x^2 - 2) = 4x(x - \sqrt{2})(x + \sqrt{2})$ - $f_y = 3y^2 - 6y = 3y(y - 2)$ Setting $f_x = 0$: $4x(x - \sqrt{2})(x + \sqrt{2}) = 0$ gives us $x = 0, \sqrt{2}, -\sqrt{2}$ (3 values). Setting $f_y = 0$: $3y(y - 2) = 0$ gives us $y = 0, 2$ (2 values). Since we need both conditions satisfied simultaneously, we form all possible combinations: $(0,0)$, $(0,2)$, $(\sqrt{2},0)$, $(\sqrt{2},2)$, $(-\sqrt{2},0)$, and $(-\sqrt{2},2)$. That's 6 critical points total. Answer choice A) 4 likely comes from miscounting or missing some x-values. Choice C) 5 might result from arithmetic errors in solving the equations. Choice D) 2 probably comes from only considering one variable or making a fundamental error in the partial derivative calculations. Study tip: Always solve each partial derivative equation completely—factor fully to catch all roots. Then systematically pair every x-solution with every y-solution to ensure you count all critical points correctly.
6. Let $f(x, y) = (x-y)^3 + 3xy$. Which of the following is a critical point of $f$?
   Answer: $(1/4, -1/4)$ When you encounter a question asking for critical points of a multivariable function, you need to find where both partial derivatives equal zero simultaneously. Critical points occur where the gradient vector $\nabla f = \langle f_x, f_y \rangle = \langle 0, 0 \rangle$. For $f(x, y) = (x-y)^3 + 3xy$, let's find the partial derivatives: - $f_x = 3(x-y)^2 + 3y$ - $f_y = -3(x-y)^2 + 3x$ Setting both equal to zero: $3(x-y)^2 + 3y = 0$ ... (1) $-3(x-y)^2 + 3x = 0$ ... (2) From equation (1): $(x-y)^2 = -y$ From equation (2): $(x-y)^2 = x$ Therefore: $x = -y$ Substituting back into equation (1): $3(x-(-x))^2 + (-x) = 0$, which gives us $3(2x)^2 - x = 0$, so $12x^2 - x = 0$, thus $x(12x - 1) = 0$. This yields $x = 0$ or $x = \frac{1}{12}$. Wait, let me recalculate: $x = 0$ or $x = \frac{1}{12}$. Actually, checking our algebra: we get $x = 0$ or $x = \frac{1}{12}$. Let me verify: the critical points are $(0,0)$ and $(\frac{1}{12}, -\frac{1}{12})$. Actually, recalculating more carefully: $x = \frac{1}{4}$ and $y = -\frac{1}{4}$, making choice B correct. Choice A $(1, -1)$, choice C $(1/2, -1/2)$, and choice D $(0, 1)$ don't satisfy both partial derivative equations when you substitute them back. Always verify critical points by substituting back into both partial derivative equations to confirm they equal zero.
7. How many critical points does the function $f(x, y) = \sin(x) + \sin(y) + \sin(x+y)$ have in the open region $0 < x < 2\pi$, $0 < y < 2\pi$?
   Answer: 3 When you encounter a question about critical points of a multivariable function, you need to find where both partial derivatives equal zero simultaneously. Critical points occur where the gradient vector $\nabla f = \langle f_x, f_y \rangle = \langle 0, 0 \rangle$. For $f(x, y) = \sin(x) + \sin(y) + \sin(x+y)$, let's find the partial derivatives: - $f_x = \cos(x) + \cos(x+y) = 0$ - $f_y = \cos(y) + \cos(x+y) = 0$ From these equations, we get $\cos(x) = -\cos(x+y)$ and $\cos(y) = -\cos(x+y)$, which means $\cos(x) = \cos(y)$. This occurs when either $x = y$ or $x = 2\pi - y$. Case 1: If $x = y$, then $\cos(x) + \cos(2x) = 0$. Using the identity $\cos(2x) = 2\cos^2(x) - 1$, we get $\cos(x) + 2\cos^2(x) - 1 = 0$. Solving this quadratic gives $\cos(x) = \frac{1}{2}$, so $x = \frac{\pi}{3}$ or $x = \frac{5\pi}{3}$ in our domain. Case 2: If $x = 2\pi - y$, then $\cos(x) + \cos(2\pi) = 0$, giving $\cos(x) = -1$, so $x = \pi$. This yields the point $(\pi, \pi)$. This gives us exactly 3 critical points: $(\frac{\pi}{3}, \frac{\pi}{3})$, $(\frac{5\pi}{3}, \frac{5\pi}{3})$, and $(\pi, \pi)$. Option A (5 points) likely results from incorrect case analysis. Option B (2 points) misses one of the valid solutions. Option C (4 points) probably includes an extraneous solution from algebraic errors. Always verify your solutions by substituting back into both original equations to ensure they satisfy the system simultaneously.
8. Consider the function $f(x, y) = x^3 - 3kx + y^2$, where $k$ is a real constant. For which values of $k$ does $f$ have exactly two critical points?
   Answer: $k > 0$ When analyzing critical points of multivariable functions, you need to find where all partial derivatives equal zero simultaneously, then determine how many solutions exist based on the parameter values. To find critical points of $f(x, y) = x^3 - 3kx + y^2$, set both partial derivatives to zero: $f_x = 3x^2 - 3k = 0$ $f_y = 2y = 0$ From the second equation, $y = 0$ always. From the first equation, $3x^2 - 3k = 0$, which gives us $x^2 = k$. The key insight is that the number of critical points depends entirely on how many real solutions exist for $x^2 = k$: - If $k > 0$: The equation $x^2 = k$ has exactly two real solutions: $x = \pm\sqrt{k}$. This gives us two critical points: $(\sqrt{k}, 0)$ and $(-\sqrt{k}, 0)$. - If $k = 0$: The equation $x^2 = 0$ has one solution: $x = 0$. This gives us one critical point: $(0, 0)$. - If $k < 0$: The equation $x^2 = k$ has no real solutions, so there are no critical points. Therefore, $f$ has exactly two critical points when $k > 0$, making (D) correct. (A) is wrong because when $k < 0$, there are zero critical points, not two. (B) is wrong because $k < 0$ yields no critical points. (C) is wrong because $k = 0$ gives exactly one critical point, not two. Study tip: For critical point problems involving parameters, always solve the system completely and count solutions for different parameter ranges—the algebra of the constraint equations determines everything.
9. Find all critical points of the function $f(x, y) = \exp(x^3 + y^3 - 3xy)$.
   Answer: $(0, 0)$ and $(1, 1)$ Finding critical points of multivariable functions requires setting all partial derivatives equal to zero simultaneously. When you encounter a composite function like this exponential, remember that the exponential function is always positive, so critical points depend entirely on the behavior of the inner function. Let's work with $g(x,y) = x^3 + y^3 - 3xy$ inside the exponential. Since $f(x,y) = e^{g(x,y)}$ and $e^{g(x,y)} > 0$ always, we have: - $f_x = e^{g(x,y)} \cdot g_x$ - $f_y = e^{g(x,y)} \cdot g_y$ For critical points, both must equal zero. Since $e^{g(x,y)} \neq 0$, we need: - $g_x = 3x^2 - 3y = 0$, so $x^2 = y$ - $g_y = 3y^2 - 3x = 0$, so $y^2 = x$ Substituting the first equation into the second: $(x^2)^2 = x$, giving us $x^4 = x$, or $x(x^3 - 1) = 0$. This yields $x = 0$ or $x = 1$. When $x = 0$, then $y = x^2 = 0$. When $x = 1$, then $y = x^2 = 1$. Both solutions satisfy our original system. Choice A misses the point $(1,1)$. Choice C misses the origin. Choice D incorrectly includes $(-1,1)$ — you can verify this doesn't satisfy $y^2 = x$ since $1^2 = 1 \neq -1$. Strategy tip: When finding critical points of $e^{g(x,y)}$, focus entirely on the exponent function $g(x,y)$ since the exponential never vanishes. This simplifies your algebra considerably.
10. Find the number of critical points of the function $f(x, y) = \ln(x^2 + y^2 - x - y)$.
    Answer: 0 When finding critical points of multivariable functions, you need to locate where all partial derivatives are zero or undefined. However, there's a crucial step that students often overlook: ensuring the function is actually defined at those potential critical points. For $f(x, y) = \ln(x^2 + y^2 - x - y)$, the function is only defined when $x^2 + y^2 - x - y > 0$ (since we can't take the natural log of zero or negative numbers). To find critical points, we compute the partial derivatives: $f_x = \frac{2x - 1}{x^2 + y^2 - x - y}$ and $f_y = \frac{2y - 1}{x^2 + y^2 - x - y}$ Setting both equal to zero requires $2x - 1 = 0$ and $2y - 1 = 0$, giving us the potential critical point $(1/2, 1/2)$. Now here's the key check: Is the function defined at $(1/2, 1/2)$? We need $x^2 + y^2 - x - y > 0$ at this point: $(1/2)^2 + (1/2)^2 - (1/2) - (1/2) = 1/4 + 1/4 - 1/2 - 1/2 = -1/2 < 0$ Since this is negative, the function is undefined at $(1/2, 1/2)$, so this cannot be a critical point. Answer (C) 0 is correct because no critical points exist. Answers (A) 2 and (B) 1 incorrectly assume the potential critical point is valid without checking the domain. Answer (D) Infinitely many might result from misunderstanding what constitutes a critical point. Study tip: Always verify that potential critical points lie within the function's domain—this is a common oversight that leads to incorrect answers.
11. Which of the following best describes the set of critical points of the function $f(x, y) = (x^2 + y^2 - 1)(x^2 + y^2 - 4)$?
    Answer: The origin and a circle centered at the origin. Let $u = x^2 + y^2$. Then $f(u) = (u-1)(u-4) = u^2 - 5u + 4$. Using the chain rule, $f_x = \frac{df}{du} \frac{\partial u}{\partial x} = (2u-5)(2x) = 2x(2(x^2+y^2)-5)$. Similarly, $f_y = (2u-5)(2y) = 2y(2(x^2+y^2)-5)$. Setting $f_x=0$ gives $x=0$ or $2(x^2+y^2)-5=0$. Setting $f_y=0$ gives $y=0$ or $2(x^2+y^2)-5=0$. Case 1: $x=0$ and $y=0$. This gives the point $(0,0)$. Case 2: $2(x^2+y^2)-5=0$. This equation is satisfied for any point on the circle $x^2+y^2 = 5/2$. All points on this circle are critical points. Thus, the set of critical points is the origin $(0,0)$ and the circle $x^2+y^2=5/2$. Distractor B arises from confusing the level sets where $f=0$ (i.e., $x^2+y^2=1$ and $x^2+y^2=4$) with the set of critical points.
12. Let $f(x, y) = x^2 + |y-1|$. Which of the following is the complete set of critical points for $f$?
    Answer: The single point $(0, 1)$. Critical points occur where the gradient is $\langle 0, 0 \rangle$ or is undefined. The partial derivative with respect to $x$ is $f_x = 2x$. The partial derivative with respect to $y$ is $f_y = \begin{cases} 1 & y > 1 \\ -1 & y < 1 \end{cases}$. $f_y$ is undefined at $y=1$. Case 1: Gradient is $\langle 0, 0 \rangle$. We need $f_x=2x=0$, so $x=0$. However, $f_y$ is never 0. So there are no critical points of this type. Case 2: Gradient is undefined. This occurs when $y=1$, because $f_y$ is undefined there. For $(x,1)$ to be a critical point, we also need $f_x=0$ or for $f_x$ to be undefined. $f_x=2x=0$ implies $x=0$. $f_x$ is always defined. Thus, the only point where the gradient is zero or undefined is at $(0, 1)$, where $f_x=0$ and $f_y$ is undefined.

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