$f(x)=\log_2(x^2-3x-4)$
Find $f^{-1}(x)$
My approach:
$y=\log_2(x^2-3x-4)$
$x=\log_2(y^2-3y-4)$
$2^x=y^2-3y-4$
$2^x+4=y(y-3)$
This is where I am stuck in my attempt on the problem.
$f(x)=\log_2(x^2-3x-4)$
Find $f^{-1}(x)$
My approach:
$y=\log_2(x^2-3x-4)$
$x=\log_2(y^2-3y-4)$
$2^x=y^2-3y-4$
$2^x+4=y(y-3)$
This is where I am stuck in my attempt on the problem.
$f$ is not injective. Inverse does not exist.