Question

For the function $f$, use composition of functions to show that $f^{-1}$ is as given. $\frac{8}{9}x + 1$, $f^{-1}(x) = \frac{9x - 9}{8}$ To show that $f^{-1}(x) = x$, find $(f \circ f^{-1})(x)$ and $(f^{-1} \circ f)(x)$ and check to see that each is x. Beginning with $(f^{-1} \circ f)(x)$. $(f^{-1} \circ f)(x) = $

          For the function $f$, use composition of functions to show that $f^{-1}$ is as given.
$\frac{8}{9}x + 1$, $f^{-1}(x) = \frac{9x - 9}{8}$
To show that $f^{-1}(x) = x$, find $(f \circ f^{-1})(x)$ and $(f^{-1} \circ f)(x)$ and check to see that each is x.
Beginning with $(f^{-1} \circ f)(x)$.
$(f^{-1} \circ f)(x) = $

For the function f, use composition of functions to show that f^-1 is as given.
(8)/(9)x + 1, f^-1(x) = (9x - 9)/(8)
To show that f^-1(x) = x, find (f ∘ f^-1)(x) and (f^-1∘ f)(x) and check to see that each is x.
Beginning with (f^-1∘ f)(x).
(f^-1∘ f)(x) =

Added by Janice W.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert K H

07/01/2021

Step 1

Step 1: Find (f^(-1)@f)(x) (f^(-1)@f)(x) = f^(-1)(f(x)) Show more…

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For the function f, use composition of functions to show that f^(-1) is as given. f(x) = (8/9)x + 1, f^(-1)(x) = (9x - 9)/8 To show that f^(-1)(x) = (9x - 9)/8, find (f^(-1)@f)(x) and (f@f^(-1))(x) and check to see that each is x. Beginning with (f^(-1)@f)(x): (f^(-1)@f)(x) = f^(-1)(f(x)) = f^(-1)((8/9)x + 1) = f^(-1)((8/9)x + 9/9) = f^(-1)((8x + 9)/9) = (9(8x + 9)/9 - 9)/8 = (72x + 81 - 9)/8 = (72x + 72)/8 = 9x + 9 Now, let's find (f@f^(-1))(x): (f@f^(-1))(x) = f(f^(-1)(x)) = f((9x - 9)/8) = (8/9)((9x - 9)/8) + 1 = x - 1 + 1 = x Therefore, both (f^(-1)@f)(x) and (f@f^(-1))(x) are equal to x, confirming that f^(-1)(x) = (9x - 9)/8.