Given a randomised square root function select the possible ways of writing the domain of the function.
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", "advice": "a) The domain of $f(x)=\\sqrt{x}$ is the set of all non-negative numbers, i.e. $\\{x\\in\\mathbb{R}:\\,x\\ge0\\}$ or in interval notation, $[0,\\infty)$. This means that the square root of any negative number is not defined.
\n\nb) For a function such as $g(t)=\\sqrt{\\simplify{t-{c[0]}}}$ we require that the square root acts on a non-negative number, that is, $\\simplify{t-{c[0]}>=0}$. Rearranging this inequality for $t$ gives $\\simplify{t>={c[0]}}$. Therefore, $\\text{dom}(g)=\\{t\\in\\mathbb{R}:\\,\\simplify{t>={c[0]}}\\}$.
\n\nc) Note, $\\sqrt[2]{\\quad}$ and $\\sqrt{\\quad}$ mean the same thing. For a function such as $h(s)=\\simplify{{c[0]}root({c[1]}s+{c[2]},2)+{c[3]}}$ we require that the square root acts on a non-negative number, that is, $\\simplify{{c[1]}s+{c[2]}>=0}$. Rearranging this inequality for $s$ gives $s>= \\simplify[fractionNumbers]{{-c[2]/c[1]}}$ $s<= \\simplify[fractionNumbers]{{-c[2]/c[1]}}$. Therefore, $\\text{dom}(h)=\\{s\\in\\mathbb{R}:\\, s\\ge \\simplify[fractionNumbers]{{-c[2]/c[1]}}\\}$ $\\text{dom}(h)=\\{s\\in\\mathbb{R}:\\, s\\leq \\simplify[fractionNumbers]{{-c[2]/c[1]}}\\}$.
\n\nd) For a function such as $\\simplify{{out}({inp})}=\\sqrt{\\simplify{{inp}^2-{a+b}{inp}+{a*b}}}$ we require that the square root acts on a non-negative number, that is, $\\simplify{{inp}^2-{a+b}{inp}+{a*b}>=0}$. Notice $\\simplify{{inp}^2-{a+b}{inp}+{a*b}=({inp}-{a})({inp}-{b})}$, which is non-negative when $\\simplify{({inp}-{a})}$ and $\\simplify{({inp}-{b})}$ are both negative, or both non-negative. That is, we require $\\simplify{{inp}<={a}}$ or $\\simplify{{inp}>={b}}$. Therefore, $\\text{dom}(\\simplify{{out}})=\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}<={a}} \\text{ or } \\simplify{{inp}>={b}}\\}$.
\n\ne) For a function such as $\\simplify{{out}({inp})=root(-{inp}^2+{d}{inp},2)}$ we require that the square root acts on a non-negative number, that is, $\\simplify{-{inp}^2+{d}{inp}>=0}$. Notice $\\simplify{-{inp}^2+{d}{inp}={inp}({d}-{inp})}$, which is non-negative when $\\simplify{{inp}}$ and $\\simplify{({d}-{inp})}$ are both negative, or both non-negative. That is, we require $\\simplify{0<={inp}<={d}}$. Therefore, $\\text{dom}(\\simplify{{out}})=\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{0<={inp}<={d}}\\}$.
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\n[[0]] $\\in \\text{dom}(f)$
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