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IgorA88
@IgorA88
January 2020
1
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(PUC-PR) o domínio da função y= 1/ [√(32 - 4^-x)]
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hcsmalves
A função estará definida, se:
32 - 4⁻ˣ > 0
32 - (1/4)ˣ > 0
-(1/4)ˣ > - 32
(1/4)ˣ < 32
(1/2)²ˣ < 2⁵
Como a base está entre 0 e 1, então:
0 < 2x < 5 (dividindo por 2)
0 < x < 5/2
D = {x ∈ IR/ 0 < x < 5/2}
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32 - 4⁻ˣ > 0
32 - (1/4)ˣ > 0
-(1/4)ˣ > - 32
(1/4)ˣ < 32
(1/2)²ˣ < 2⁵
Como a base está entre 0 e 1, então:
0 < 2x < 5 (dividindo por 2)
0 < x < 5/2
D = {x ∈ IR/ 0 < x < 5/2}