Mathematics
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):

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