Let \(f:\mathbb{R}\rightarrow\mathbb{R}\) defined by \(f(x)=5x+4\) then \(f^{-1}(x)\) is

\(\frac{x-4}{5}\) \(\frac{x+4}{5}\) \(\frac{x-4}{4}\) f is non invertible

\(\frac{x-4}{5}\)

\(f(x)=5x+4\) so $y=5x+4⇒y-4=5x$ $⇒x=\frac{y-4}{5}$ $f^{-1}(x)=\frac{x-4}{5}$