SOLUTION: Is it possible for x = 3 to be in the domains of the functions. q(x)=2x square root of 2 divided by x-3 and T(x)= square root this is inside the sign 2-x. I did q(3)=2(3) wit

Algebra ->  Algebra  -> Linear-equations -> SOLUTION: Is it possible for x = 3 to be in the domains of the functions. q(x)=2x square root of 2 divided by x-3 and T(x)= square root this is inside the sign 2-x. I did q(3)=2(3) wit      Log On

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Question 215294: Is it possible for x = 3 to be in the domains of the functions.
q(x)=2x square root of 2 divided by x-3 and
T(x)= square root this is inside the sign 2-x.
I did q(3)=2(3) with square root of 2 divided by 3-3 and For T(3)= square root that is inside the sign is 2-3.
Why or why not? I think q(3)=2(3)is yes because after working it out on my calculater I came out with 3 and the second one no because it came out undefined.
What are the domains of q(x) and T(x) ? (3,3),(0,0)
Please explain what I did wrong if possible. Thank You
Answer by Theo(3458) About Me  (Show Source):
You can put this solution on YOUR website!
your equation is q%28x%29+=+2%2A%28sqrt%282%2F%28x-3%29%29%29
A graph of this equation is shown below.
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C2%2A%28sqrt%282%2F%28x-3%29%29%29%29
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at x = 3, (x-3) in the denominator causes a division by zero which causes the value to be undefined at that point.
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At x < 3, (x-3) causes the denominator to be negative which causes the expression within the square root sign to be negative which results in the function not having real values at that point.
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The domain of the function q%28x%29+=+2%2A%28sqrt%282%2F%28x-3%29%29%29 is all value of x > 3.
The range of the function is all values of y > 0.
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y will never be equal to 0 no matter how large x gets because the numerator in the fraction underneath the square root sign is a constant. It will approach 0 but never touch it.
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The same graph but carrying x out a lot further shows this to be true.
graph%28800%2C300%2C-100%2C10000%2C-1%2C1%2C2%2A%28sqrt%282%2F%28x-3%29%29%29%29
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I didn't understand the second part of your question.
Maybe answering the first part helped you with the second part?
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