Question
Please check my answers and let me know if I did something wrong. Thank you!
Find the partial derivative of x, and the partial derivative of y, then the partial derivative of x (1,-1), and the partal derivative of y (1,-1).
f(x,y) = x^4 y^2 -x
here is what I got
@f/@x = 3x^3 y^2 -1
@f / @y = x^4 2y
@f / @x = 3(1)^3 (-1)^2 -1 = 2
@f / @y = (1)^4 2(-1) = -2
f(x,y) = e^(2x+y)
@f/@x = 2e^(2x+y)
@f / @y = e^(y+2x)
@f / @x = 2e
@f / @y = e
Locate and classify the critical points the functions.
g(x,y) = -x^2 -2xy + y^2 + x -4y
g(x) = -2x -2y +1
g(y) = -2x + 2y -4
set them equal to zero?
g(x) = -2x -2y +1 = 0
g(y) = -2x + 2y -4 = 0
Here is what I got.
(-3/4, 5/4)
f(x,y) xe^y
f(x) = e^y
f(y) = e^(y) x
x=0
Find the partial derivative of x, and the partial derivative of y, then the partial derivative of x (1,-1), and the partal derivative of y (1,-1).
f(x,y) = x^4 y^2 -x
here is what I got
@f/@x = 3x^3 y^2 -1
@f / @y = x^4 2y
@f / @x = 3(1)^3 (-1)^2 -1 = 2
@f / @y = (1)^4 2(-1) = -2
f(x,y) = e^(2x+y)
@f/@x = 2e^(2x+y)
@f / @y = e^(y+2x)
@f / @x = 2e
@f / @y = e
Locate and classify the critical points the functions.
g(x,y) = -x^2 -2xy + y^2 + x -4y
g(x) = -2x -2y +1
g(y) = -2x + 2y -4
set them equal to zero?
g(x) = -2x -2y +1 = 0
g(y) = -2x + 2y -4 = 0
Here is what I got.
(-3/4, 5/4)
f(x,y) xe^y
f(x) = e^y
f(y) = e^(y) x
x=0
Answers
Looks good, except for the last one. It has no critical points, since Fx is never zero.
Note that the critical points are where both Fx and Fy are zero.
Note that the critical points are where both Fx and Fy are zero.
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