Question 683045
Correct the denominator cannot be 0,
(x-4)(x-1) = 0
x-4=0
x = 4
x-1=0
x = 1
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But also, there should not be a negative under the square root
{{{sqrt(x-6) >= 0}}}
{{{sqrt(x-6)^2 >= 0^2}}}
{{{x-6 >= 0}}}
{{{x >= 6}}}
...
Since 1 and 4 are less than 6, they don't need a separate mention.
...
The domain is simply all real numbers greater than or equal to 6.
[6, {{{infinity}}})
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