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
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) $$\displaystyle{g{{\left({f{{\left(-{2}\right)}}}\right)}}}$$

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 $$\displaystyle{\left(-\infty,{0}\right]}\cup{\left[{2},\infty\right)}$$