Question

Compose the following function and state the domain of the composed function. f(x)= 1/x-1 g(x)= sqrt(1-x^2) a) g(f(-2))

Composite functions
ANSWERED
asked 2021-01-10
Compose the following function and state the domain of the composed function.
\(\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{x}}-{1}\)
\(\displaystyle{g{{\left({x}\right)}}}=\sqrt{{{1}-{x}^{{2}}}}\)
a) \(\displaystyle{g{{\left({f{{\left(-{2}\right)}}}\right)}}}\)

Answers (1)

2021-01-11
To find g(f(−2)), substitute -2 for x in the function \(\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{x}−{1}}}\) and then \(\displaystyle−\frac{{1}}{{3}}\) for x in the function \(\displaystyle{g{{\left({x}\right)}}}=\sqrt{{{1}-{x}^{{2}}}}\), and simplify.
\(\displaystyle{g{{\left({f{{\left(−{2}\right)}}}\right)}}}={g{{\left(\frac{{1}}{{-{2}-{1}}}\right)}}}\)
\(\displaystyle=\sqrt{{{1}-{\left(-\frac{{1}}{{3}}\right)}^{{2}}}}\)
\(\displaystyle=\sqrt{{{1}-\frac{{1}}{{9}}}}\)
\(\displaystyle=\sqrt{{\frac{{{9}-{1}}}{{9}}}}\)
\(\displaystyle=\frac{\sqrt{{8}}}{{3}}\)
\(\displaystyle=\frac{{{2}\sqrt{{2}}}}{{3}}\)
Therefore, \(\displaystyle{g{{\left({f{{\left(-{2}\right)}}}\right)}}}=\frac{{{2}\sqrt{{2}}}}{{3}}\)
To obtain the composite function \(\displaystyle{\left({g}\circ{f}\right)}{\left({x}\right)}\), substitute \(\displaystyle{f{{\left({x}\right)}}}=\frac{{1}}{{{x}−{1}}}\) and \(\displaystyle{g{{\left({x}\right)}}}=\sqrt{{{1}-{x}^{{2}}}}\) in the definition of composite function \(\displaystyle{\left({g}\circ{f}\right)}{\left({x}\right)}={g}{\left[{f{{\left({x}\right)}}}\right]}\), and simplify.
\(\displaystyle{\left({g}\circ{f}\right)}{\left({x}\right)}={g}{\left[{f{{\left({x}\right)}}}\right]}\)
\(\displaystyle={g}{\left|\frac{{1}}{{{x}-{1}}}\right|}\)
\(\displaystyle=\sqrt{{{1}-{\left(\frac{{1}}{{{x}-{1}}}\right)}^{{2}}}}\)
\(\displaystyle=\sqrt{{\frac{{{\left({x}-{1}\right)}^{{2}}-{1}}}{{\left({x}-{1}\right)}^{{2}}}}}\)
\(\displaystyle={\left(\frac{\sqrt{{{x}^{{2}}-{2}{x}+{1}-{1}}}}{{{x}-{1}}}\right.}\)
\(\displaystyle=\frac{{\sqrt{{{x}{\left({x}-{2}\right)}}}}}{{{x}-{1}}}\)
Therefore, \(\displaystyle{g}{\left[{f{{\left({x}\right)}}}\right]}=\frac{{\sqrt{{{x}{\left({x}-{2}\right)}}}}}{{{x}-{1}}}\)
To determine the domain of the composite function \(\displaystyle{\left({g}\circ{f}\right)}{\left({x}\right)}\), use the definition of the composite function \(\displaystyle{g}{\left[{f{{\left({x}\right)}}}\right]}=\frac{{\sqrt{{{x}{\left({x}-{2}\right)}}}}}{{{x}-{1}}}\) and identify the values of x, for which the denominator is not equal to zero and radicand is positive.
\(\displaystyle{x}\le{0}\)
\(\displaystyle{x}-{2}\ge{0}\)
\(\displaystyle{x}\ge{2}\)
\(\displaystyle{x}-{1}\ne{0}\)
\(\displaystyle{x}\ne{1}\)
Hence, the domain of the composite function \(\displaystyle{g}{\left[{f{{\left({x}\right)}}}\right]}=\frac{{\sqrt{{{x}{\left({x}-{2}\right)}}}}}{{{x}-{1}}}\) is \(\displaystyle{\left(-\infty,{0}\right]}\cup{\left[{2},\infty\right)}\)
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