t Given
[tex]f(x)=\sqrt{x^2-1}[/tex]
Find
inverse of f(x)
domain and range of function and its inverse
Explanation
Let y = f(x)
[tex]y=\sqrt{x^2-1}[/tex]
replace all x with y and y with x
[tex]x=\sqrt{y^2-1}[/tex]
now solve for y
[tex]\begin{gathered} \sqrt{y^2-1}=x \\ y^2-1=x^2 \\ y^2=x^2+1 \\ y=\pm\sqrt{x^2+1} \end{gathered}[/tex]
so, the inverse is
[tex]\pm\sqrt{x^2+1}[/tex]
domain of f(x) is
[tex]x\epsilon R:x\leq-1\text{ or x}\ge1[/tex]
range of f(x) is
[tex]y\epsilon R:y\ge0[/tex]
domain of inverse is
[tex]R[/tex]
range of inverse function is
[tex]y\epsilon R:y\ge1[/tex]
