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DanaMadalina
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va rog ajutati-ma
1. consideram functia f:R->R, f(x)=4. Determinati valoarea raportului f(0)/2
2.fie functia f:R->R, f(x)=-2x+√6. Aratati ca a=f(√3)-f(√2)/√2-√3 ∈ N
3.Fie f:{-3,0,3,6} ->R,f(x)=1-4/3x. Calculati f(6)-f(-3)

Răspuns :

1. f(0)/2=4/2=2
2. a=[f(√3)-f(√2)]/(√2-√3)=(-2√3+√6+2√2-√6)/(√2-√3)=2(√2-√3)/(√2-√3)=2
3. f(6)-f(-3)=1-4/18-1-4/9=-6/9=-2/3

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