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

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(B) For the function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ to be defined, both square root expressions must be non-negative.Step $1$: For $\sqrt{x-1}$, we

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$[1, 6]$

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(B) For the function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ to be defined, both square root expressions must be non-negative.Step $1$: For $\sqrt{x-1}$, we require $x-1 \ge 0$, which implies $x \ge 1$.Step $2$: For $\sqrt{6-x}$, we require $6-x \ge 0$, which implies $x \le 6$.Step $3$: The domain is the intersection of these two conditions: $x \ge 1$ and $x \le 6$.Step $4$: Combining these, we get $1 \le x \le 6$, which in interval notation is $[1, 6]$.

The domain of the function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ is...

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