1/(x+4)^2

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Solution for 1/(x+4)^2 equation: 


D( x )

(x+4)^2 = 0

(x+4)^2 = 0

(x+4)^2 = 0

x+4 = 0 // - 4

x = -4

x in (-oo:-4) U (-4:+oo)

1/((x+4)^2) = 0

x belongs to the empty set

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