Answer: [tex]\dfrac{x^2+1}{2}[/tex]
Step-by-step explanation:
Given
[tex]f(x)=\sqrt{2x-1}[/tex]
We can write it as
[tex]\Rightarrow y=\sqrt{2x-1}[/tex]
Express x in terms of y
[tex]\Rightarrow y^2=2x-1\\\\\Rightarrow x=\dfrac{y^2+1}{2}[/tex]
Replace y be x to get the inverse
[tex]\Rightarrow f^{-1}(x)=\dfrac{x^2+1}{2}[/tex]
To prove, it is inverse of f(x). [tex]f(f^{-1}(x))=x[/tex]
[tex]\Rightarrow f(f^{-1}(x))=\sqrt{2\times \dfrac{x^2+1}{2}-1}\\\\\Rightarrow f(f^{-1}(x))=\sqrt{x^2+1-1}\\\\\Rightarrow f(f^{-1}(x))=x[/tex]
So, they are inverse of each other.
