Chanel Hunter

2023-02-27

The function f is defined as $f\left(x\right)=\frac{x}{x-1}$, how do you find $f\left(f\left(x\right)\right)$?

Caiden Mitchell

Beginner2023-02-28Added 3 answers

Given: $f\left(x\right)=\frac{x}{x-1}$

Substitute f(x) for every x

$f\left(f\left(x\right)\right)=\frac{\frac{x}{x-1}}{\left(\frac{x}{x-1}\right)-1}$

Multiply 1 to the numerator and denominator to create the following: $\frac{x-1}{x-1}$

$f\left(f\left(x\right)\right)=\frac{\frac{x}{x-1}}{\left(\frac{x}{x-1}\right)-1}\frac{x-1}{x-1}$

$f\left(f\left(x\right)\right)=\frac{x}{x-x+1}$

$f\left(f\left(x\right)\right)=\frac{x}{1}$

$f\left(f\left(x\right)\right)=x$

This means that $f\left(x\right)=\frac{x}{x-1}$ is its own inverse.

Substitute f(x) for every x

$f\left(f\left(x\right)\right)=\frac{\frac{x}{x-1}}{\left(\frac{x}{x-1}\right)-1}$

Multiply 1 to the numerator and denominator to create the following: $\frac{x-1}{x-1}$

$f\left(f\left(x\right)\right)=\frac{\frac{x}{x-1}}{\left(\frac{x}{x-1}\right)-1}\frac{x-1}{x-1}$

$f\left(f\left(x\right)\right)=\frac{x}{x-x+1}$

$f\left(f\left(x\right)\right)=\frac{x}{1}$

$f\left(f\left(x\right)\right)=x$

This means that $f\left(x\right)=\frac{x}{x-1}$ is its own inverse.

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