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12 Nov 2019
Can you help step 3 question?
Examine the function for relative extrema and saddle points. g(x, y)x2-y* - 8x - 7y Step 1 Find the first partial derivatives of gx, y)x2 - y2 - 8x Ty with respect to x and y. 2-8 -2y-7 Observe that both the partial derivatives existexist for all x and y. So according to the definition of critical points, the only critical points are where the first partial derivatives are equal to 0. Step 2 To locate the critical points, set gx(x. y)0 and gylx, y) 0. -2y-7 = 0 Solve the equations to find the values of x and y 4,-3.5 This is the one critical point of the function. Step 3 of the function g ecu Findthe val g(x, y)2 - y2 - 8x - 7y relative extrema of the function g occurs at 4, Find the value -11.73x
Can you help step 3 question?
Examine the function for relative extrema and saddle points. g(x, y)x2-y* - 8x - 7y Step 1 Find the first partial derivatives of gx, y)x2 - y2 - 8x Ty with respect to x and y. 2-8 -2y-7 Observe that both the partial derivatives existexist for all x and y. So according to the definition of critical points, the only critical points are where the first partial derivatives are equal to 0. Step 2 To locate the critical points, set gx(x. y)0 and gylx, y) 0. -2y-7 = 0 Solve the equations to find the values of x and y 4,-3.5 This is the one critical point of the function. Step 3 of the function g ecu Findthe val g(x, y)2 - y2 - 8x - 7y relative extrema of the function g occurs at 4, Find the value -11.73x
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Tod ThielLv2
30 Apr 2019