Răspuns:
a)f(x)=(2x+5)/(4x+3)
Verifici da ca functia sa fie bijectiva, adica injectiva si surjectiva
i.injectivA f(x1)=f(x2)
(2x1+5)/(4x1+3)=(2x2+5)/(4x2+3)
(2x1+5)(4x2+3)=(2x2+5)(4x1+3)
8x1x2+20x2+6x1+15=8x1x2+20x1+6x2+15
20x2+6x1-20x1-6x2=0
20x2-20x1)-(6x2-6x1)=0
20(x2-x1)-6(x2-x1)=0
14(x2-x1=0=>x1=x2 => f injectiva
surjectiva
f(x)=y
y=(2x+5)/(4x+3)
y(4x+3)=2x+5
4xy+3y=2x+5
4xy-2x=5-3y
2x(2y-1)=5-3y
x=(5-3y)/2(2y-1)
Deci 2y-1≠0
y≠1/2
y∈R\{1/2}
imf=codomeniu =>
f surjectiva
Deoarece f este si injectiva si surjectiva, atunci este bijectiva
Am aratat ca y=(5-3y)/2(2y-1)
f(y)=(5-3y)/2(2y-1)
Faci schimbarea de variabila
y=x
f⁻¹(x)=(5-3x)/2(2x-1)
Revin imediat
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b)f(x)=(x-1)/(x+1)
verifici bijectivitatea
injectiva f(x1)=f(x2)
f(x!)=f(x2)
(x1-1)/(x1+1)=(x2-1)/(x2+1)
(x1-1)(x2+1)=(x2-1)(x1+1)
x1x2-x2+x1-1=x1x2-x1+x2-1
(x1-x2)=(x2-x1)
(x1-x2)-(x2-x1)=0
(x1-x2)+(x1-x2)=0
2(x1-x2)=0
x1-x2=0
x1=x2 Functie injectiva
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surjectivitatea
y=(x-1)/(x+!)
yx+y=x-1
yx-x=-y-1
x= (-y-1)/(y-1)
x=(y+1)/(y-1)
y≠1
Imf=domeniu=R\{1} functia surjectiva
Inversa
x=(Y+!)/(y-1)
f(y)=(y+1)/(y-1)
y=x
f⁻¹(x)=(x+1)/(x-1)
Explicație pas cu pas:
