vestirme4

2021-10-29

Find the inverse the function and the domain and the range of f and ${f}^{-1}$, if the function is $f\left(x\right)=2\frac{x}{x+5}$ and it's one-to-one.

Sally Cresswell

To find the inverse function, substitute $x={f}^{-1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=x$ and calculate:
$y=2\frac{x}{x+5}$
$xy+5y=2x$
$x\left(y-2\right)=-5y$
$x=\frac{-5y}{2-y}$
$x=\frac{5y}{2-y}$
${f}^{-1}\left(x\right)=\frac{5x}{2-x}$
If $f\left({f}^{-1}\left(x\right)\right)={f}^{-1}\left(f\left(x\right)\right),$
then ${f}^{-1}\left(x\right)$ is an inverse function of $f\left(x\right)$
$=f\left({f}^{-1}\left(x\right)\right)$
$=f\left(\frac{5x}{2-y}\right)$
$=\frac{2×\frac{5x}{2-y}}{\frac{5x}{2-x}+5}$
$=\frac{\frac{10x}{2-x}}{\frac{5x+10-5x}{2-x}}$
=x
The domain of $f\left(x\right)$ is range for ${f}^{-1}\left(x\right)$
And, the range of $f\left(x\right)$ is domain for ${f}^{-1}\left(x\right)$
$f\left(x\right)=\frac{2x}{x+5}$
x+5=0
x=-5
Domain $=\left(-\mathrm{\infty },-5\right)\cup \left(-5,\mathrm{\infty }\right)$
${f}^{-1}\left(x\right)=\frac{5x}{2-x}$
2-x=0
x=2
Range $=\left(-\mathrm{\infty },2\right)\cup \left(2,\mathrm{\infty }\right)$

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