f(x)=3/(x²+2x+1)+√(x+3)
f(x)=3/(x+1)²+√(x+3)
Wyznaczymy Df :
x+1≠0 ∧ x+3 ≥ 0
x≠-1 ∧ x ≥ -3
x∈<-3,-1)∪(-1,∞)
Df=<-3,-1)∪(-1,∞)
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f(x)=3/(x²+2x+1)+√(x+3)
f(x)=3/(x+1)²+√(x+3)
Wyznaczymy Df :
x+1≠0 ∧ x+3 ≥ 0
x≠-1 ∧ x ≥ -3
x∈<-3,-1)∪(-1,∞)
Df=<-3,-1)∪(-1,∞)