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

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(C) For the function $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$ to be defined,the expression inside the logarithm must be positive,and the express

Vedclass · Accepted Answer

$[4, 6]$

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(C) For the function $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$ to be defined,the expression inside the logarithm must be positive,and the expressions inside the square roots must be non-negative.$1$. For $\sqrt {x - 4}$ to be defined,$x - 4 \ge 0$,which implies $x \ge 4$.$2$. For $\sqrt {6 - x}$ to be defined,$6 - x \ge 0$,which implies $x \le 6$.$3$. The sum $\sqrt {x - 4} + \sqrt {6 - x}$ must be strictly greater than $0$. Since both square root terms are non-negative,their sum is always $\ge 0$. The sum is $0$ only if $x-4=0$ and $6-x=0$ simultaneously,which is impossible ($x=4$ and $x=6$). Thus,the sum is always positive for $x \in [4, 6]$.Combining these conditions,we get $4 \le x \le 6$.Therefore,the domain is $[4, 6]$.

The domain of the function $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$ is

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