(2x^3/2-3x^-3/2)^2

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Solution for (2x^3/2-3x^-3/2)^2 equation:


D( x )

x = 0

x = 0

x = 0

x in (-oo:0) U (0:+oo)

((2*x^3)/2-((3*x^-3)/2))^2 = 0

((2*x^3)/2+(-3/2)*x^-3)^2 = 0

(x^3-3/2*x^-3)^2 = 0

x^3-3/2*x^-3 = 0

x^-3*(x^6-3/2) = 0

1*x^6 = 3/2 // : 1

x^6 = 3/2

x^6 = 3/2 // ^ 1/6

abs(x) = (3/2)^(1/6)

x = (3/2)^(1/6) or x = -(3/2)^(1/6)

x^-3*(x-(3/2)^(1/6))*(x+(3/2)^(1/6)) = 0

(x^-3)^2*(x-(3/2)^(1/6))^2*(x+(3/2)^(1/6))^2 = 0

( x+(3/2)^(1/6) )

x+(3/2)^(1/6) = 0 // - (3/2)^(1/6)

x = -(3/2)^(1/6)

( x-(3/2)^(1/6) )

x-(3/2)^(1/6) = 0 // + (3/2)^(1/6)

x = (3/2)^(1/6)

( x^-3 )

1*x^-3 = 0 // : 1

x^-3 = 0

x należy do O

x in { -(3/2)^(1/6), -(3/2)^(1/6), (3/2)^(1/6), (3/2)^(1/6) }

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