Find the Domain and Range of f(x)=√(x²-16)

Find the Domain and Range of $f(x)=\sqrt{x^2-16}$

Question: Find the domain and range of the real valued function: $$ f(x)=\sqrt{x^2-16} $$

Solution

For a square root function, the expression inside the root must be non-negative.

Therefore, $$ x^2-16\ge0 $$

$$ (x-4)(x+4)\ge0 $$

This is true when $$ x\le-4 \quad \text{or} \quad x\ge4 $$

Hence, the domain is: $$ (-\infty,-4]\cup[4,\infty) $$

Since square root values are always non-negative, $$ f(x)\ge0 $$

Minimum value occurs at $$ x=\pm4 $$

$$ f(\pm4)=\sqrt{16-16}=0 $$

As $|x|$ increases, $$ x^2-16 $$ increases without bound.

Therefore, $$ f(x)\to\infty $$

Hence, the range is: $$ [0,\infty) $$

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