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Beach Google Slides, Using CAS to find Max, Min, Saddle Points of Multivariable .. 9:41In this video we go over how to use your TI-Nspire CAS to find and classify the critical points of a multivariable function.YouTube · turksvids · Feb 23, 202010 key moments in this video Black History Google Slides Template Free, Critical/Saddle point calculator for f(x,y). Critical/Saddle point calculator for f(x,y). 2 min read. Critical/Saddle point . It has a global maximum point and a local extreme maxima point at X.Questions & answersStack ExchangeQuestionCalculate the partial derivative, local minima/maxima, and saddle points.Answer · 1 votePartial derivatives are calculated by regarding the function as a function in only one argument and considering the other variables as constants.So you obtain fx(x,y)=ddx(yx−2x−1−2/y)=y+2x−2.Saddle points and local maxima/minima are always at places where both derivatives vanish simultaneously. To decide which is which, you have to look at the second derivatives:• fx(x0,y0)>0,fy(x0,y0)>0: (x0,y0) is local minimum• fx(x0,y0)<0,fy(x0,y0)<0: (x0,y0) is local maximum• fx(x0,y0)>0,fy(x0,y0)<0 orfx(x0,y0)<0,fy(x0,y0>0: (x0,y0) is a saddle point.MoreCheggQuestionFind the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.) f(x,y)=(x−y)(25−xy) local maximumAnswer · 0 votesTo find the local maximum and minimum values and saddle points of the function we first calculate the.MoreCheggQuestion5-22 Find the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function.16. f(x,y)=x2+y4+2xyAnswer · 0 votesGiven the function f(x,y)=x ^2 +y^ 4 +2xy calculate the partial derivatives w.r.t. x and y so (delf)/(delx)=del/(delx)(x ^2 +y^ 4 +2xy) =2x+2yMoreNumeradeQuestionFind the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function. f(x, y)=2 x^2-8 x y+y^4-4 y^3Answer · 2 votesSo we're given f of x of y is equal to 2 x, squared minus 8 x y plus y to the fourth minus 4 y cube, so we're encouraged to use our graphing calculator, but what we're going to find is a maximum or minimum. My graphing calculator give me something like this, then i zoomed out- and i actually got something more like this, so it must have been those wires. I was seeing that it wants to put on there because it is a plane, so we're going to find what this minimum value is. So we're going to differentiate with respect of x, we get 4 x minus 8 y and then, if with respect of y, we get negative 8 x plus 4 y cube minus 12 y squared we're going to set each of these equal to 0 and we're going To solve so this first 1 we're going to solve for x, so 4 x, minus 8 y equals 04 x, equals 8 y x equals to y, so we're gonna plug that into this equation. So negative 8 times 2 y plus 4 y cube minus 12 y. Squared is equal to 0, so negative 16 y plus 4 y cube minus 12 y squared equals 0 and we're going to div…MoreNumeradeQuestionText: 5, 6, 2, 8, 2, 10, 11, 12, 13, 14, 15, 16, 12, 18, 19, 20, 21, and 22. Find the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function.Answer:113 Answer: 15, 16. f(1, y) = 2x^2 + y^4 + 2xyAnswer · 3 votesThe first step is to find the critical points by taking the partial derivative first with respect to x, f prime of x, and this will be 3 x squared times y. We don't do anything to the y. This would be 72 x and we have nothing else, because this is treated as a constant. Then we need to take the derivative with respect to y, so this would be x, cubed, and this would be minus 8, because this is treated as a constant. Now we set them equal to 0, we set them equal to 0, let's start with the bottom fraction for bottom function, because we don't have any is the only x, so we can solve for x right away. We said it equal to 0- let's put it in blue, so we can keep track of this so x, cubed minus 8 equals 0. The 8 goes as a positive, so x, cubed is equal to 8 x is equal to cubic root of 8 is equal to 2 point. Now we go back to our other function so 3 and replace x, squared for a 2 point so 2 to the power of 2 y plus 72 times 2 equals 0. This will be 3 times 4. This will be 12 y and this is equal…MoreCourse HeroQuestion. Find the local maximum and minimum values and saddle point(s).Answer · 1 voteCALCULUS Image transcriptions f ( x ,y ) = 2y cos x; 05 x5 2TT for ( x ,y )= - 2y sin (xx) Fy ( xx, 4) = 2 cos ( xx ) for = 0, fy = 0 - 24 sin (x ) = 0, 2 cos ( 2 ) = 0 y=0, sin (x ) = 0; x = 0, TT, 2 TT cos ( x ) = 0 x = II 3TT All points to be considered are : " (/2, 0) ( 315, 0), (0, 0), ( TT, ), ( 27 , 0 ) fxx = -24 cos (a)fyy = 0 i foxy = - 2 sin 2 Dat (o, 0) , fxx = 0, fyy= ofxy= 0 2 at (TO fxx = 0, foy = 0 , fay = - 2 D= faculfyy - (focy ) = 0 (0) - (-2)? = - 440 Local maximum values ONE Local minimum values ONE Saddle points ( x ,y ) = 371MoreCourse HeroQuestionFind the local maximum and minimum values and saddle point(s) of.Answer · 0 votesAnswer- Local maximum value=4 Local minimum value=-4 Saddle points=(0, 1) and (0, -1) Image transcriptions Ans - f (20 , 4 ) = 20- 6 + 604 2 @ The first step is to find all the first-order partial derivatives : - f x ( 2 14 ) = 6 ( 20 2 + 4 2 - 1 ) 2 ay f (ary ) = a (2x3 - 62 + 6 24 2 ) Fy ( # 14 ) = 12 4 2) Next, solve the system, at = 0 of = 0 6 ( x 2 + 4 2- 1 ) = 6 2 4 12ay = 0. 2+ 4 2-1 =0 or 12 my = 0. The system has following seal solutions- (any) = (-110), ( x , 4 ) = (0, -17 , ( x,4)= (0.1) (,y ) = ( 1 10) 3) Now, Let's toy to classify them find all the second order partial derivatives? 2 2 2x 2 ( 2 5 - 6 0ey 2 - 6 29 ) = 12C 2 2 rehe ( zx 3 - 6 aly 2 - 620 ) = 1 2 4 2 2 242 ( 250 3 - Grey 2-6 x ) = 12 20 Define the expression D = 2of a2f a2 f 2 chee ay 2 = (124. 1220 ) - ( 12 4 ) = 144 ( 249 - 1 = ] = 144 ( 2 - 4 ) ( 2+y )D = 144 ( 2 - 4 ) ( Hty ) since, D ( - 1 10) = 14470 8 2 2 -42 ( 20 46x4 - 64 ) is at (xly) = (-1 10) is -12 < 0 It can be stated that (-110 ) is a relativ…MoreBrainly.comQuestionFind the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpoint that reveals all the important aspects of the function. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.) f(x,y)=2x 3 −6x+6xy 2 local maximum value(s) local minimum value(s) saddle point(s) (x,y)=(0,1),(0,−1)Answer · 0 votesThe function f(x,y)=2x^3−6x+6xy^2 has local maximum value(s) at (0,1) and (0,−1), local minimum value(s) at (0,1) and (0,−1), and saddle point(s) at (0,1) and (0,−1).To determine the local maximum and minimum values, as well as the saddle points of the given function f(x,y), we first need to find its critical points. Critical points occur when the partial derivatives with respect to x and y are equal to zero or do not exist. In this case, the function has two critical points: (0,1) and (0,−1).Next, we use the second derivative test to classify these critical points. For a function of two variables like this one, we need to compute the determinant of the Hessian matrix. If the determinant is positive, it's a local minimum; if it's negative, it's a local maximum; and if it's zero, the test is inconclusive.At both critical points, the Hessian determinant turns out to be zero, which means the second derivative test is inconclusive. Therefore, we cannot definitively determine whether these …MoreBartleby.comQuestionFind the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the function with a domain and viewpointthat reveals all the important aspects of the function. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.)f(x, y) = x3 + y3 – 3x2 – 9y2 – 9xlocal maximum value(s)local minimum value(s)saddle point(s)(х, у) %3Answer · 0 votesStep 1 .MoreWyzantQuestionFind the local max and min values and saddle points. (Partial derivatives)Answer · 0 votesf(x,y) = xy/exp[(x2+y2)/2]Find the gradient:∂f/∂x = {y exp[(x2+y2)/2] - x2y exp[(x2+y2)/2]}/exp(x2+y2)= y(1-x2)/exp[(x2+y2)/2]∂f/∂y = x(1-y2)/exp[(x2+y2)/2]So the gradient is 0 if x(1+y)(1-y)=0 and y(1+x)(1-x)=0We get (0,0), (1,1), (1,-1), (-1,1), and (-1,-1).We test these with a Hessian.∂2f/∂x2 = {-2xy exp[(x2+y2)/2] - xy(1-x2)exp[(x2+y2)/2]}/exp(x2+y2)= xy(x2-3)/exp[(x2+y2)/2]∂2f/∂y2 = 2xy(y2-3)/exp[(x2+y2)/2]∂2f/∂x∂y = {(1-y2)exp[(x2+y2)/2] - x2(1-y2)exp[(x2+y2)/2]}/exp(x2+y2)= (1-x2)(1-y2)/exp[(x2+y2)/2]det[Hf(x,y)] = [4x2y2(x2-3)(y2-3) - (1-x2)(1-y2)]/exp(x2+y2)=Plug the points in:At (0,0), ∂2f/∂x2 = 0.At (1,1) and (-1,-1), ∂2f/∂x2 = -2/e < 0.At (-1,1) and (1,-1), ∂2f/∂x2 = 2/e > 0.At (0,0), det[Hf(x,y)] = -1/e2 < 0At (1,1), (-1,-1), (-1,1), and (1,-1), det[Hf(x,y)] = 36/e2 > 0Saddle points: f = 0 at (0,0).Maxima: f = 1/e at (1,1) and (-1,-1)Minima: f = -1/e at (-1,1) and (1,-1)MoreGauthmathQuestionQUESTION 13 Find all the local maxima, local minima, and saddle points of the function. fx,y=x2+20x+y2+2y-8 f10,1=295 , local maximum f-10,-1=-109 , local minimum f10,-1=291 , saddle point f-10,1=-105 , saddle pointAnswer · 146 votesAnswer: .More Feedback Bop Or Flop Google Slides, Find the local maximum and minimum values and saddle .. Question: Find the local maximum and minimum values and saddle point(s) of the function. You are encouraged to use a calculator or computer to graph the .1 answer  ·  Top answer: To find the local maximum and minimum values and saddle points of the function we first calculate the. Brown Aesthetic Google Slides, Find Extreme and Saddle Points - Derivative Calculator. Extreme Points Calculator is an online tool to get the maxima and minima of derivative function. Make calculation of saddle points easy with extrema calculator.VideosYou will see more English now.14:35Local Extrema, Critical Points, & Saddle Points of Multivariable .YouTube · The Organic Chemistry Tutor14 minutes, 35 secondsNov 6, 20199 key moments in this videoFrom 00:00IntroFrom 02:39Example #1From 04:19Second Partial Derivatives and Mixed Partial DerivativesFrom 06:09Evaluate the point at the functionFrom 07:44Partial derivative with respect to y11:23Local extrema and saddle points of a multivariable function .YouTube · Krista King11 minutes, 23 secondsMay 29, 20149 key moments in this videoFeedbackView allPeople also askYou will see more English now.How do you determine min max or saddle point?Is a saddle point a local maximum?How do you find local maxima and minima and saddle point?What is the formula for saddle point?FeedbackWeb results Camping Google Slides Theme, Calculus III - Relative Minimums and Maximums. 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