sqrt{x + 2sqrt{x - 1}} + sqrt{x - 2sqrt{x - 1 | vedclass

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Question

(A) Let $f(x) = \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}}$.Since the term under the square root must be non-negative,$x - 1 \ge 0$,so $x \ge

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$2$,if $1 \le x \le 2$

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(A) Let $f(x) = \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}}$.Since the term under the square root must be non-negative,$x - 1 \ge 0$,so $x \ge 1$.We can rewrite the expression as $\sqrt{(\sqrt{x-1})^2 + 1 + 2\sqrt{x-1}} + \sqrt{(\sqrt{x-1})^2 + 1 - 2\sqrt{x-1}}$.This simplifies to $\sqrt{(\sqrt{x-1} + 1)^2} + \sqrt{(\sqrt{x-1} - 1)^2}$.This is equal to $|\sqrt{x-1} + 1| + |\sqrt{x-1} - 1|$.Since $\sqrt{x-1} + 1$ is always positive for $x \ge 1$,we have $(\sqrt{x-1} + 1) + |\sqrt{x-1} - 1|$.Case $1$: If $1 \le x \le 2$,then $0 \le \sqrt{x-1} \le 1$,so $|\sqrt{x-1} - 1| = 1 - \sqrt{x-1}$.Thus,$f(x) = \sqrt{x-1} + 1 + 1 - \sqrt{x-1} = 2$.Case $2$: If $x > 2$,then $\sqrt{x-1} > 1$,so $|\sqrt{x-1} - 1| = \sqrt{x-1} - 1$.Thus,$f(x) = \sqrt{x-1} + 1 + \sqrt{x-1} - 1 = 2\sqrt{x-1}$.

$\sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} = $

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