Question

Arătați ca următoarele funcții sunt inversabile si determinați invesele lor

Answer 1

a)

f(x₁)=f(x₂)=>(x₁+3)/(x₁+1)=(x₂+3)/(x₂+1)<=>x₁x₂+x₁+3x₂+3=x₁x₂+3x₁+x₂+3 |-x₁x₂-3=>x₁+3x₂=3x₁+x₂<=>x₁-3x₁=x₂-3x₂<=>-2x₁=-2x₂ |*(-1/2)=>x₁=x₂ (A)

f(x₁)=f(x₂)=>x₁=x₂=>f injectiva (1)

∀y∈(1,3),∃x∈(0,∞) a.i.y=f(x)=>y=(x+3)/(x+1)<=>yx+y=x+3<=>yx-x=3-y<=>x(y-1)=3-y=>x=(3-y)/(y-1)

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

Din (1) si (2)=> f bijectiva=>f inversabila , f⁻¹:(1,3)->(0,∞)

y=f(x)=>y=(x+3)/(x+1)<=>yx+y=x+3<=>yx-x=3-y<=>x(y-1)=3-y=>x=(3-y)/(y-1)=>f⁻¹(x)=(3-x)/(x-1)

f⁻¹:(1,3)->(0,∞) , f⁻¹(x)=(3-x)/(x-1)

b)

f(x₁)=f(x₂)=>x₁²+1=x₂²+1 |-1=>x₁²=x₂² |√=>x₁=x₂ (A)

f(x₁)=f(x₂)=>x₁=x₂=>f injectiva (1)

∀y∈(1,∞),∃x∈(0,∞) a.i.y=f(x)=>y=x²+1=>-x²=1-y=>x²=y-1=>x=√(y-1)

y∈(1,∞)=>y>1 |-1=>y-1>0 |√=>√(y-1)>0=>x>0=>x∈(0;∞) (A)=>f surjectiva (2)

Din (1) si (2)=> f bijectiva=>f inversabila , f⁻¹:(1,∞)->(0,∞)

y=f(x)=>y=x²+1=>-x²=1-y=>x²=y-1=>x=√(y-1)=>f⁻¹(x)=√(x-1)

f⁻¹:(1,∞)->(0,∞) ,f⁻¹(x)=√(x-1)