$$
f(x, y)=\frac{x^{2}}{2}-x^{4}+\cos (x)-y^{2}+2 y
$$

What are all the critical points of $f$ ?
Choose 1 answer:
(A) $(-0.511 \ldots, 1)$
(B) $(0,1)$
(C) $(0.511 \ldots, 1)$
(D) There are no critical points.

Question

$$
f(x, y)=\frac{x^{2}}{2}-x^{4}+\cos (x)-y^{2}+2 y
$$

What are all the critical points of $f$ ?
Choose 1 answer:
(A) $(-0.511 \ldots, 1)$
(B) $(0,1)$
(C) $(0.511 \ldots, 1)$
(D) There are no critical points.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the critical points of the function $f(x, y) = \frac{x^2}{2} - x^4 + \cos(x) - y^2 + 2y$, we need to find where the gradient of $f$ is zero. This means solving $ abla f = \vec{0}$. ‖ ‖ step 2 ⋮ The partial derivative with respect to $x$ is $f_x(x, y) = x - 4x^3 + \sin(x)$. Setting this equal to zero gives us the equation $x - 4x^3 + \sin(x) = 0$. ‖ ‖ step 3 ⋮ The partial derivative with respect to $y$ is $f_y(x, y) = -2y + 2$. Setting this equal to zero gives us $-2y + 2 = 0$, which simplifies to $y = 1$. ‖ ‖ step 4 ⋮ Substituting $y = 1$ into $f_x(x, y) = 0$, we need to solve $x - 4x^3 + \sin(x) = 0$ for $x$. This is a transcendental equation and may have multiple solutions. ‖ ‖ step 5 ⋮ The asksia-ll calculator has provided the solutions to the equation $x - 4x^3 + \sin(x) = 0$ as $x \approx -0.511...$ and $x \approx 0.511...$. ‖ ‖ step 6 ⋮ Therefore, the critical points of the function are approximately $(-0.511..., 1)$ and $(0.511..., 1)$. ‖ ∻[question 1] Answer∻ ***A and C*** ∻Key Concept∻ ⚹Critical Points of a Multivariable Function⚹ ∻Explanation∻ ⚹Critical points occur where the gradient of the function is zero. For a function $f(x, y)$, this means solving $f_x(x, y) = 0$ and $f_y(x, y) = 0$ simultaneously.⚹◊What is the formula to determine the critical points of a multivariable function?⍭ Generate me a similar question◊

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