The domain of the function f(x) = sqrt{x - 1} | vedclass

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(C) For the function $f(x) = \sqrt{x - 1} + \sqrt{3 - x}$ to be defined, the expressions under the square roots must be non-negative.Step $1$: For $\s

Vedclass · Accepted Answer

$[1, 3]$

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(C) For the function $f(x) = \sqrt{x - 1} + \sqrt{3 - x}$ to be defined, the expressions under the square roots must be non-negative.Step $1$: For $\sqrt{x - 1}$, we require $x - 1 \ge 0$, which implies $x \ge 1$.Step $2$: For $\sqrt{3 - x}$, we require $3 - x \ge 0$, which implies $x \le 3$.Step $3$: The domain is the intersection of these two conditions: $x \ge 1$ and $x \le 3$.Step $4$: Combining these, we get $1 \le x \le 3$, which is the interval $[1, 3]$.

The domain of the function $f(x) = \sqrt{x - 1} + \sqrt{3 - x}$ is ...

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