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

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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Mitchel Aguirre
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
$$\displaystyle{g}{o}{f{{\left({x}\right)}}}=\sqrt{{{x}^{{2}}-{5}}}$$
$$\displaystyle\sqrt{{{x}^{{2}}-{5}}}\ge{0}$$
$$\displaystyle{x}^{{2}}-{5}\ge{0}$$
$$\displaystyle{x}\ge\sqrt{{5}}$$
$$\displaystyle-\sqrt{{5}}\le{x}\le\sqrt{{5}}$$
$$\displaystyle{x}\in{\left[-\sqrt{{5}},\sqrt{{5}}\right]}$$