1/(16-x^2)^(0.5)

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Solution for 1/(16-x^2)^(0.5) equation: 


D( x )

(16-x^2)^0.5 = 0

16-x^2 < 0

(16-x^2)^0.5 = 0

(16-x^2)^0.5 = 0

-1*x^2 = -16 // : -1

x^2 = 16

x^2 = 16 // ^ 1/2

abs(x) = 4

x = 4 or x = -4

16-x^2 < 0

16-x^2 < 0

-1*x^2 < -16

-1*x^2 < -16 // * -1

1*x^2 > 16

1*x^2 > 16 // : 1

x^2 > 16/1

x^2 > 16

x^2 > 16 // ^ 1/2

abs(x) > 4

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

x in (-4:4)

1/((16-x^2)^0.5) = 0

x belongs to the empty set

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