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Let f be a re...

Question

Let $f$ be a real-valued invertible function such that $f (\frac{2 x - 3}{x - 2}) = 5 x - 2, x \neq 2.$ Then the value of $f^{- 1} (13)$ is

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Solution

$f (\frac{2 x - 3}{x - 2}) = 5 x - 2 \Rightarrow f^{- 1} (5 x - 2) = \frac{2 x - 3}{x - 2}$
Let $5 x - 2 = 13,$ then $x = 3$
Hence $f^{- 1} (13) = \frac{2 (3) - 3}{3 - 2} = 3$

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