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

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Question

(B) The function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ is defined if and only if the expressions under the square roots are non-negative. For $\sqrt{x-1}$

Vedclass · Accepted Answer

$[1, 6]$

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(B) The function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ is defined if and only if the expressions under the square roots are non-negative. For $\sqrt{x-1}$ to be defined,$x - 1 \geq 0$,which implies $x \geq 1$. For $\sqrt{6-x}$ to be defined,$6 - x \geq 0$,which implies $x \leq 6$. Combining these two conditions,we get $1 \leq x \leq 6$. Therefore,the domain of $f(x)$ is $[1, 6]$.

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

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