## Calculus: Early Transcendentals (2nd Edition)

a. $f^{-1}(x) = 4x-4$ b. $f(f^{-1}(x)) = (4x-4)/4 + 1$ $f(f^{-1}(x)) = x - 1 + 1$ $f(f^{-1}(x)) = x$ $f^{-1}(f(x)) = 4(x/4 - 1) + 4$ $f^{-1}(f(x)) = x - 4 + 4$ $f^{-1}(f(x)) = x$
a. To find the inverse of a function $f(x)$, switch $f(x)$ with $x$ and replace $f(x)$ with $f^{-1}(x)$ $f(x) = x/4 + 1$ $x = f^{-1}(x)/4 + 1$ $x - 1 = f^{-1}(x)/4$ $4x-4 = f^{-1}(x)$ $f^{-1}(x) = 4x-4$ b. $f(f^{-1}(x)) = (4x-4)/4 + 1$ $f(f^{-1}(x)) = x - 1 + 1$ $f(f^{-1}(x)) = x$ $f^{-1}(f(x)) = 4(x/4 - 1) + 4$ $f^{-1}(f(x)) = x - 4 + 4$ $f^{-1}(f(x)) = x$