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Find the domain of the function f(x) = sqrt(sqrt(x^2 - 16) + 3).

 May 3, 2022
 #1
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If there is a square root, the expression inside must be nonnegative.

 

As a result, \(x^2 - 16 \geq 0\). You can solve this to get the domain of f(x).

 

Remarks: You do not need to consider \(\sqrt{x^2 - 16} + 3 \geq 0\) because \(\sqrt{x^2 - 16} + 3 \geq 0 + 3 = 3 > 0\) if sqrt(x^2 - 16) is defined.

 May 3, 2022

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