How do you find the domain of 1/(sqrt(17-t))?

1 Answer
Somebody N.
Mar 28, 2018

{t in RR | -oo< t < 17}

Explanation:

1/sqrt(17-t)

For real numbers the radicand ( value inside the radical ) must be greater than zero. We can't allow zero, because this would be undefined, ( division by zero ).

:.

17-t>0

t<17

So domain is:

(-oo,17)

or in set notation:

{t in RR | -oo< t < 17}

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