Find the domain and range of f(x) = \(\sqrt{16-x}\).

1.	Find the domain and range of f(x) = \(\sqrt{16-x}\).
Answer» if you mean f(x)=¹⁶√-x then, as the root is an even integer, the unknown under the root must take a negative value to reverse the negative sign of (-x) into a positive value. SO the domain will take the non-positive portion of the x-axis -> (-∞,0] the range will be [0,∞) the non-negative portion of the y-axis hope that helps f(x) = \(\sqrt{16-x}\) \(\because\) Domain of square root function \(\sqrt{g(x)}\) is g(x) \(\geq 0\). \(\therefore\) For Domain of given function f(x) 96 - x \(\geq 0\) ⇒ x \(\leq\) 16 ⇒ x \(\in\) (-\(\infty\), 16].

Answer»

if you mean f(x)=¹⁶√-x then, as the root is an even integer, the unknown under the root must take a negative value to reverse the negative sign of (-x) into a positive value. SO the domain will take the non-positive portion of the x-axis -> (-∞,0]

the range will be [0,∞) the non-negative portion of the y-axis

hope that helps

f(x) = \(\sqrt{16-x}\)

\(\because\) Domain of square root function \(\sqrt{g(x)}\) is g(x) \(\geq 0\).

\(\therefore\) For Domain of given function f(x)

96 - x \(\geq 0\)

⇒ x \(\leq\) 16

⇒ x \(\in\) (-\(\infty\), 16].

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