Answer:
a.Domain of f=(1.099,[tex]\infty)[/tex]
b.[tex]f^{-1}(x)=ln(e^x+3)[/tex]
Step-by-step explanation:
Let [tex]y=f(x)=ln(e^x-3)[/tex]
We know that domain of ln x is greater than zero
[tex]e^x-3>0[/tex]
Adding 3 on both sides of inequality
[tex]e^x-3+3>0+3[/tex]
[tex]e^x>3[/tex]
Taking on both sides of inequality
[tex]lne^x>ln 3[/tex]
[tex]x>ln 3[/tex]=1.099
By using [tex]lne^x=x[/tex]
Domain of f=(1.099,[tex]\infty)[/tex]
Let [tex]y=f^{-1}(x)=ln(e^x-3)[/tex]
[tex]e^y=e^x-3[/tex]
By using property [tex]lnx=y\implies x=e^y[/tex]
[tex]e^x=e^y+3[/tex]
Taking ln on both sides of equality '
[tex]lne^x=ln(e^y+3)[/tex]
[tex]x=ln(e^y+3)[/tex]
Replace x by y and y by x
[tex]y=ln(e^x+3)[/tex]
Substitute y=[tex]f^{-1}(x)[/tex]
[tex]f^{-1}(x)=ln(e^x+3)[/tex]
