The domain of the function sqrt{log(x^2 - 6x | vedclass

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(C) The function $f(x) = \sqrt{\log(x^2 - 6x + 6)}$ is defined when the expression inside the square root is non-negative,i.e.,$\log(x^2 - 6x + 6) \ge

Vedclass · Accepted Answer

$( - \infty, 1 ] \cup [ 5, \infty )$

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(C) The function $f(x) = \sqrt{\log(x^2 - 6x + 6)}$ is defined when the expression inside the square root is non-negative,i.e.,$\log(x^2 - 6x + 6) \ge 0$.Since $\log(y) \ge 0$ implies $y \ge 1$ (assuming base $10$ or $e$),we have:$x^2 - 6x + 6 \ge 1$$x^2 - 6x + 5 \ge 0$Factoring the quadratic expression:$(x - 5)(x - 1) \ge 0$To solve this inequality,we find the critical points $x = 1$ and $x = 5$. Testing the intervals $(-\infty, 1]$,$(1, 5)$,and $[5, \infty)$,we find that the expression is non-negative in the intervals $(-\infty, 1]$ and $[5, \infty)$.Additionally,we must ensure the argument of the logarithm is positive: $x^2 - 6x + 6 > 0$. Since $x^2 - 6x + 6 \ge 1$ is already satisfied,the condition $x^2 - 6x + 6 > 0$ is automatically satisfied.Therefore,the domain is $( - \infty, 1 ] \cup [ 5, \infty )$.

The domain of the function $\sqrt{\log(x^2 - 6x + 6)}$ is

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