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

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

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There is a single real number that is not in the domain of the function 

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\\[f(x)=\\frac{1}{x}.\\]

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That number is [[0]]. 

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There are [[0]] real numbers that are not in the domain of {rat} these are [[1]].

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Note: If the numbers were $-2,1$ and $4$ you would enter set(-2,1,4)

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$\\mathbb{R}$

", "

$\\{x\\in \\mathbb{R}:\\,\\var{a+b}<x<\\var{a*b}\\}$

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$\\{x\\in \\mathbb{R}:\\,x\\ne \\var{c[0]},\\var{c[1]}\\}$

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$\\{x\\in \\mathbb{R}:\\,x\\ne \\var{a},\\var{b}\\}$

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$\\mathbb{R}\\setminus \\{\\var{a},\\var{b}\\}$

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$\\mathbb{R}\\setminus \\{\\var{c[0]},\\var{c[1]}\\}$

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$\\{x\\in \\mathbb{R}:\\,x\\ne \\var{-a},\\var{-b}\\}$

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Which of the following represents the domain of \\[\\simplify{g(x)=(x^2-{c[0]+c[1]}x+{c[0]*c[1]})/(x^2-{a+b}x+{a*b})}?\\]

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$\\{r\\in\\mathbb{R}:\\,r\\ne \\var{c[0]}\\}$

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$\\mathbb{R}$

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Which of the following is the domain of

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\\[h(r)=\\simplify{(r-{c[0]})/(r-{c[0]})}?\\]

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\n

Division is defined for all real numbers except zero. So when a function involves division (such as a rational function), any input that will result in a division by zero is not allowed. Rational functions are simply a fraction/division of polynomials, so the only real numbers that are not in the domain of a rational function are the roots of the polynomial in the denominator. Sometimes, (for example part c above) you will need to factorise the denominator to determine its roots.

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d) Even though 'something divided by itself is 1' division by zero is still undefined. So the domain of $h$ is not all of $\\mathbb{R}$, the domain does not include the number $\\var{c[0]}$. In other words, $h(\\var{c[0]})$ is undefined but for all other $r$, $h(r)=1$.

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