Is the square root of 16 - x^2 a one-to-one function?
Added by Angel R.
Step 1
Since we are taking the square root of a quantity, the radicand (16-x^2) must be non-negative. Therefore, we have: 16-x^2 ≥ 0 x^2 ≤ 16 -4 ≤ x ≤ 4 So the domain of the function is [-4,4]. Show more…
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