Domain of the function f(x) = sqrt{2 - 2x - x | vedclass

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Question

(B) For the function $f(x) = \sqrt{2 - 2x - x^2}$ to be defined,the expression under the square root must be non-negative.Thus,$2 - 2x - x^2 \ge 0$.Mu

Vedclass · Accepted Answer

$-1 - \sqrt{3} \le x \le -1 + \sqrt{3}$

Vedclass · Answer

(B) For the function $f(x) = \sqrt{2 - 2x - x^2}$ to be defined,the expression under the square root must be non-negative.Thus,$2 - 2x - x^2 \ge 0$.Multiplying by $-1$,we get $x^2 + 2x - 2 \le 0$.To solve this inequality,we complete the square:$x^2 + 2x + 1 - 1 - 2 \le 0$$(x + 1)^2 - 3 \le 0$$(x + 1)^2 \le 3$Taking the square root on both sides,we get:$-\sqrt{3} \le x + 1 \le \sqrt{3}$Subtracting $1$ from all parts:$-1 - \sqrt{3} \le x \le -1 + \sqrt{3}$.Therefore,the domain is $[-1 - \sqrt{3}, -1 + \sqrt{3}]$.

Domain of the function $f(x) = \sqrt{2 - 2x - x^2}$ is

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