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

vestirme4 2021-10-29 Answered
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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Expert Answer

Sally Cresswell
Answered 2021-10-30 Author has 3772 answers
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 \(\displaystyle={\left(-\infty,{2}\right)}\cup{\left({2},\infty\right)}\)
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