What is the domain of #sqrt(x^2-9)#?

Answer 1

The domain of this expression is #{x|x < -3# or #x > 3}#.

To find restrictions to the domain of a square root expression, you must determine which values of the variable will make the radicand have a negative value.

#x^2 - 9 < 0# #x^2 - 9 + 9 < 0 + 9# #x^2 < 9# #sqrtx^2 < sqrt9# #x < +-3#

This actually tells us that #x# cannot be between #-3# and #3#, inclusive. Therefore #x# must be less than #-3# or greater than #3#.

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Answer 2

Ariana Alvarez

The domain of the function sqrt(x^2 - 9) includes all real numbers for which the expression inside the square root is non-negative.

For x^2 - 9 to be non-negative, we have: x^2 - 9 ≥ 0

Solving this inequality: x^2 ≥ 9 x ≥ ±3

Therefore, the domain of sqrt(x^2 - 9) is all real numbers greater than or equal to -3 and less than or equal to 3, or in interval notation: Domain: [-3, 3]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.