The domain of the function f(x) = sqrt{x - x^ | vedclass

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Question

(D) For the function $f(x) = \sqrt{x - x^2} + \sqrt{4 + x} + \sqrt{4 - x}$ to be defined,the expressions under the square roots must be non-negative.$

Vedclass · Accepted Answer

$[0, 1]$

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(D) For the function $f(x) = \sqrt{x - x^2} + \sqrt{4 + x} + \sqrt{4 - x}$ to be defined,the expressions under the square roots must be non-negative.$1$. For $\sqrt{4 + x}$,we require $4 + x \ge 0$,which implies $x \ge -4$.$2$. For $\sqrt{4 - x}$,we require $4 - x \ge 0$,which implies $x \le 4$.$3$. For $\sqrt{x - x^2}$,we require $x - x^2 \ge 0$,which is $x(1 - x) \ge 0$. This inequality holds when $0 \le x \le 1$.To find the domain,we take the intersection of these intervals:$x \in [-4, \infty) \cap (- \infty, 4] \cap [0, 1]$.The intersection of these sets is $[0, 1]$.Therefore,the domain of the function is $[0, 1]$.

The domain of the function $f(x) = \sqrt{x - x^2} + \sqrt{4 + x} + \sqrt{4 - x}$ is

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