0=-1/(3y^2-1)

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Solution for 0=-1/(3y^2-1) equation:


D( y )

3*y^2-1 = 0

3*y^2-1 = 0

3*y^2-1 = 0

3*y^2 = 1 // : 3

y^2 = 1/3

y^2 = 1/3 // ^ 1/2

abs(y) = (1/3)^(1/2)

y = (1/3)^(1/2) or y = -(1/3)^(1/2)

y in (-oo:-(1/3)^(1/2)) U (-(1/3)^(1/2):(1/3)^(1/2)) U ((1/3)^(1/2):+oo)

0 = -1/(3*y^2-1) // + -1/(3*y^2-1)

0-(-1/(3*y^2-1)) = 0

(3*y^2-1)^-1 = 0

1/(3*y^2-1) = 0

y belongs to the empty set

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