Suppose f(x)=x^2 and g(x)=sqrt(x-5), to find: 1. (fog)(x) 2. Domai

tricotasu 2021-10-14 Answered
Suppose \(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}{\quad\text{and}\quad}{g{{\left({x}\right)}}}=\sqrt{{{x}-{5}}}\), to find:
1. (fog)(x)
2. Domain of (fog)(x)
3. (gof)(x)
4. Domain of (gof)(x)

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Expert Answer

Mitchel Aguirre
Answered 2021-10-15 Author has 17519 answers
1. \(\displaystyle{f}{o}{g{{\left({x}\right)}}}={f{{\left({g{{\left({x}\right)}}}\right)}}}\)
\(\displaystyle={f{{\left(\sqrt{{{x}-{5}}}\right)}}}\)
\(\displaystyle={\left(\sqrt{{{x}-{5}}}\right)}^{{2}}\)
=x-5
2. \(\displaystyle{g{{\left({x}\right)}}}=\sqrt{{{x}-{5}}}\)
\(\displaystyle{g{{\left({x}\right)}}}{>}{0}\) where \(\displaystyle{x}\in{\left[{5},\infty\right)}\)
3. \(\displaystyle{g}{o}{f{{\left({x}\right)}}}={g{{\left({f{{\left({x}\right)}}}\right)}}}\)
\(\displaystyle={g{{\left({x}^{{2}}\right)}}}\)
\(\displaystyle=\sqrt{{{x}^{{2}}-{5}}}\)
4. \(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}\) is true for all real numbers
Except
\(\displaystyle{g}{o}{f{{\left({x}\right)}}}=\sqrt{{{x}^{{2}}-{5}}}\)
\(\displaystyle\sqrt{{{x}^{{2}}-{5}}}\ge{0}\)
\(\displaystyle{x}^{{2}}-{5}\ge{0}\)
\(\displaystyle{x}\ge\sqrt{{5}}\)
\(\displaystyle-\sqrt{{5}}\le{x}\le\sqrt{{5}}\)
\(\displaystyle{x}\in{\left[-\sqrt{{5}},\sqrt{{5}}\right]}\)
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