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Given the real functions below, you should be able to determine their domains. 

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Give an example of a real number that is in the domain of the function

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\\[f(x)=\\sqrt{x}.\\]

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 [[0]] $\\in \\text{dom}(f)$

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$t\\ge\\var{c[0]}$

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$t> \\var{c[0]}$

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$t\\le\\var{c[0]}$

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$t< \\var{c[0]}$

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$t>0$

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$t\\ne0$

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Given the function \\[g(t)=\\sqrt{\\simplify{t-{c[0]}}},\\] what do we require of $t$ so that $g(t)$ is defined?

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}<={a}} \\text{ or } \\simplify{{inp}>={b}}\\}$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{a}<={inp} <={b}}\\}$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}>={a*b}}\\}$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}>=0}\\}$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{min([a+b,a*b])}<={inp} <={max([a+b,a*b])}}\\}$

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Given the function \\[\\simplify{{out}({inp})}=\\sqrt{\\simplify{{inp}^2-{a+b}{inp}+{a*b}}},\\] which of the following represents the domain of $\\simplify{{out}}$?

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}<=0} \\text{ or } \\simplify{{inp}>={d}}\\}$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{0<={inp} <={d}}\\}$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}>={d}}\\}$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}>=0}\\}$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{-d}<={inp} <={d}}\\}$

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Given the function \\[\\simplify{{out}({inp})=root(-{inp}^2+{d}{inp},2)},\\] which of the following represents the domain of $\\simplify{{out}}$?

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Given a randomised square root function select the possible ways of writing the domain of the function.

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base

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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. 

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b) 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, $\\text{dom}(g)=\\{t\\in\\mathbb{R}:\\,\\simplify{t>={c[0]}}\\}$.

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c) 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, $\\text{dom}(\\simplify{{out}})=\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}<={a}} \\text{ or } \\simplify{{inp}>={b}}\\}$.

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d) 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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", "name": "Maria's copy of Domain of a square root function", "preamble": {"css": "", "js": ""}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}