To do this, we can switch $x$ and $y$ and solve for $y$:
$y = 8x - 2 \Rightarrow x = 8y - 2$
Now, solve for $y$:
$x + 2 = 8y \Rightarrow y = \frac{x + 2}{8}$
So, $f^{-1}(x) = \frac{x + 2}{8}$.
Now, we can find $f(f^{-1}(9))$:
$f(f^{-1}(9)) = f\left(\frac{9 +
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