$f(x)=\log_2(x^2-3x-4)$
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.
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Apply the formula for quadratic equations and check which of the solutions satisfy $x=\log_2(y^2-3y-4)$ – Peter Nov 09 '16 at 17:03
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@Peter I don't quite understand why we should use the quadratic formula in order to find the inverse function, could you explain a bit more? – Hiro Nov 09 '16 at 17:08
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1You want to solve $y^2-3y-(2^x+4)=0$ for $y$ – Peter Nov 09 '16 at 17:12
