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OpenStudy (anonymous):

what would the domain of x/2-x be?

OpenStudy (anonymous):

did you mean, x/(2-x) or (x/2) - x

OpenStudy (anonymous):

is this \[\frac{x}{x-2}\]? if so all real numbers except 2

OpenStudy (anonymous):

because you still cannot divide by zero

OpenStudy (anonymous):

oh so do you only look at the denominator to figure out the domain?

OpenStudy (anonymous):

for this one.... \[k(x)=\sqrt{x/2-x}\]

OpenStudy (anonymous):

Domain really only considers two major elements: denominators of fractions cannot be zero (divide by 0 problem), and even radicals (such as square roots, 4th roots, etc) cannot be taken of negative values. For the problem of x/(2-x), the answer of all reals except 2 is correct. For the radical problem, you must also consider that the radicand must be greater than or equal to zero, or solve this inequality: \[x/2 - x \ge 0\]

OpenStudy (anonymous):

\[x \le -2\]

OpenStudy (anonymous):

Eliza27: Not correct for the radical problem proposed by desireeee11. Try x = 0 in the radical expression...it will not cause a domain violation. (k(0) = 0) Perhaps an algebra error?

OpenStudy (anonymous):

yes you only look at the denominator if you have a rational function

OpenStudy (anonymous):

Oh yeah I though the 0 was a 1 thanks

OpenStudy (anonymous):

if you have a radical inside must \[\geq 0\]

OpenStudy (anonymous):

"radicand" is the corect word

OpenStudy (anonymous):

but if it is really \[k(x)=\frac{x}{x-2}\]\] then solution is more complicated

OpenStudy (anonymous):

i mean \[k(x)=\sqrt{\frac{x}{x-2}}\]

OpenStudy (anonymous):

then you have to make sure that \[\frac{x}{x-2}\geq 0\]

OpenStudy (anonymous):

solution would be \[(\infty,0]\cup (2,\infty)\]

OpenStudy (anonymous):

thank you

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