f(x,y)=2x^4 - x^2 +3y^2

partial differentiation wrt x and equating to 0 gives x=0, +1/4, -1/4

partial differentiation wrt y and equating to 0 gives y=0

so the critical points for f(x,y) will be (0,0), (1/4,0) and (-1/4,0)?

Question

f(x,y)=2x^4 - x^2 +3y^2

partial differentiation wrt x and equating to 0 gives x=0, +1/4, -1/4

partial differentiation wrt y and equating to 0 gives y=0

so the critical points for f(x,y) will be (0,0), (1/4,0) and (-1/4,0)?

anonymous · Answer

my basic aint strong T_T

amistre64 · Answer

those should be critical points to test out, yes

anonymous · Answer

owh k. just wanted to confirm the critical points. sometimes the critical points are weird for certain question.