15.
Let x1ā,x2āāR and f(x)=ā3x+4.
If f(x1ā)=f(x2ā)=>ā3x1ā+4=ā3x2ā+4
=>ā3x1ā=ā3x2ā=>x1ā=x2ā.
If x1āī =x2ā=>f(x1ā)ī =f(x2ā).
Then
f(x1ā)=f(x2ā)<=>x1ā=x2ā,x1ā,x2āāR The function f(x)=ā3x+4 is one-to-one.
āyāR,y=ā3x+4,āx=ā31āy+34ā,xāR The function f(x)=ā3x+4 is onto.
The function f(x)=ā3x+4 is one-to-one and is onto.
Therefore the function f(x)=ā3x+4 is a bijection. Therefore the function f(x)=ā3x+4 is invertibele
f(x)=ā3x+4 Replace f(x) with y
y=ā3x+4 Switch x and y
x=ā3y+4 Solve for y
y=ā31āx+34ā Replace y with fā1(x)
fā1(x)=ā31āx+34ā
16.
f(x)=x+2x+1ā Domain: (āā,ā2)āŖ(ā2,ā)
Replace f(x) with y
y=x+2x+1ā Switch x and y
x=y+2y+1ā Solve for y
xy+2x=y+1
y=āxā12xā1ā Replace y with fā1(x)
fā1(x)=āxā12xā1ā
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