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intervalo √1−x
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Solution
f(x)≥0
+1
Interval Notation
[0,∞)
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Solution steps
Solve by:
One step at a time
L−1{x3−5(x−3)2}
Tirar a fração parcial de x3−5(x−3)2:x+6+27x−3+22(x−3)2
=L−1{x+6+27x−3+22(x−3)2}
Usar a propriedade de linearidade da transformada inversa de Laplace: Para as funções f(s),g(s) e constantes a,b:L−1{a·f(s)+b·g(s)}=a·L−1{f(s)}+b·L−1{g(s)}