what would the domain of x/2-x be?
did you mean, x/(2-x) or (x/2) - x
is this \[\frac{x}{x-2}\]? if so all real numbers except 2
because you still cannot divide by zero
oh so do you only look at the denominator to figure out the domain?
for this one.... \[k(x)=\sqrt{x/2-x}\]
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: \[x/2 - x \ge 0\]
\[x \le -2\]
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?
yes you only look at the denominator if you have a rational function
Oh yeah I though the 0 was a 1 thanks
if you have a radical inside must \[\geq 0\]
"radicand" is the corect word
but if it is really \[k(x)=\frac{x}{x-2}\]\] then solution is more complicated
i mean \[k(x)=\sqrt{\frac{x}{x-2}}\]
then you have to make sure that \[\frac{x}{x-2}\geq 0\]
solution would be \[(\infty,0]\cup (2,\infty)\]
thank you
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