Question

The one-to-one function $f$ is defined below. $f(x) = \frac{x-2}{5x+4}$ Find $f^{-1}(x)$, where $f^{-1}$ is the inverse of $f$. Also state the domain and range of $f^{-1}$ in interval notation. $f^{-1}(x) = \frac{2(2x+1)}{1-5x}$ Domain of $f^{-1}$ : Range of $f^{-1}$ :

          The one-to-one function $f$ is defined below.
$f(x) = \frac{x-2}{5x+4}$
Find $f^{-1}(x)$, where $f^{-1}$ is the inverse of $f$.
Also state the domain and range of $f^{-1}$ in interval notation.
$f^{-1}(x) = \frac{2(2x+1)}{1-5x}$
Domain of $f^{-1}$ : 
Range of $f^{-1}$ :

The one-to-one function f is defined below.
f(x) = (x-2)/(5x+4)
Find f^-1(x), where f^-1 is the inverse of f.
Also state the domain and range of f^-1 in interval notation.
f^-1(x) = (2(2x+1))/(1-5x)
Domain of f^-1 :
Range of f^-1 :

Added by Francisca B.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Suman Saurav Thakur

04/08/2022

Step 1

So, y = (x-2)/(5x+4) Show more…

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The one-to-one function f is defined below. f(x)=(x-2)/(5x+4) Find f^(-1)(x), where f^(-1) is the inverse of f. Also state the domain and range of f^(-1) in interval notation. The one-to-one function f is defined below x-2 5x+4 Find f- (), wheref-1 is the inverse of f. Also state the domain and range of f in interval notation. (-2(2x+1) 1-5x 00 (0,0) [0,0] (0,0] [g,0) Domain off: 8 -8 Range off: 5