# Find the inverse the function and the domain and the range of f and f^(-1)

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

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Sally Cresswell
To find the inverse function, substitute $$\displaystyle{x}={f}^{{-{1}}}{\quad\text{and}\quad}{y}={x}$$ and calculate:
$$\displaystyle{y}={2}\frac{{x}}{{{x}+{5}}}$$
$$\displaystyle{x}{y}+{5}{y}={2}{x}$$
$$\displaystyle{x}{\left({y}-{2}\right)}=-{5}{y}$$
$$\displaystyle{x}=\frac{{-{5}{y}}}{{{2}-{y}}}$$
$$\displaystyle{x}=\frac{{{5}{y}}}{{{2}-{y}}}$$
$$\displaystyle{{f}^{{-{1}}}{\left({x}\right)}}=\frac{{{5}{x}}}{{{2}-{x}}}$$
If $$\displaystyle{f{{\left({{f}^{{-{1}}}{\left({x}\right)}}\right)}}}={{f}^{{-{1}}}{\left({f{{\left({x}\right)}}}\right)}},$$
then $$\displaystyle{{f}^{{-{1}}}{\left({x}\right)}}$$ is an inverse function of $$\displaystyle{f{{\left({x}\right)}}}$$
$$\displaystyle={f{{\left({{f}^{{-{1}}}{\left({x}\right)}}\right)}}}$$
$$\displaystyle={f{{\left(\frac{{{5}{x}}}{{{2}-{y}}}\right)}}}$$
$$\displaystyle=\frac{{{2}\times\frac{{{5}{x}}}{{{2}-{y}}}}}{{\frac{{{5}{x}}}{{{2}-{x}}}+{5}}}$$
$$\displaystyle=\frac{{\frac{{{10}{x}}}{{{2}-{x}}}}}{{\frac{{{5}{x}+{10}-{5}{x}}}{{{2}-{x}}}}}$$
=x
The domain of $$\displaystyle{f{{\left({x}\right)}}}$$ is range for $$\displaystyle{{f}^{{-{1}}}{\left({x}\right)}}$$
And, the range of $$\displaystyle{f{{\left({x}\right)}}}$$ is domain for $$\displaystyle{{f}^{{-{1}}}{\left({x}\right)}}$$
$$\displaystyle{f{{\left({x}\right)}}}=\frac{{{2}{x}}}{{{x}+{5}}}$$
x+5=0
x=-5
Domain $$\displaystyle={\left(-\infty,-{5}\right)}\cup{\left(-{5},\infty\right)}$$
$$\displaystyle{{f}^{{-{1}}}{\left({x}\right)}}=\frac{{{5}{x}}}{{{2}-{x}}}$$
2-x=0
x=2
Range $$\displaystyle={\left(-\infty,{2}\right)}\cup{\left({2},\infty\right)}$$