Question

Este aceasta functie bijectiva? Cum?

Answer 1

f:R->(1,+∞)

fie a,b∈R

f(a)=f(b)<=>4ᵃ+2ᵃ+1=4ᵇ+2ᵇ+1 |-1=>2²ᵃ+2ᵃ=2²ᵇ+2ᵇ<=>2ᵃ(2+1)=2ᵇ(2+1) |:(2+1)=>2ᵃ=2ᵇ=>(bijectivitatea functiei exponentiale):a=b=>f injectiva (1)

∀y∈(1,+∞) ∃x∈R a.i. y=f(x)

y=4ˣ+2ˣ+1<=>y=2²ˣ+2ˣ+1<=>2ˣ(2+1)=y-1<=>2ˣ=(y-1)/3=>x=log₂(y-1)/3<=>log₂2ˣ=log₂(y-1)/3=>(bijectivitatea functiei logaritmice):x=(y-1)/3

y∈(1,+∞)=>y>1 |-1=>y-1>0|:3=>(y-1)/3>0=>x>0=>x∈R (A)=>f surjectiva  (2)

Din (1) si (2)=>f bijectiva