The domain of the function f(x) = sqrt{2 - se | vedclass

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Question

(C) For the function $f(x) = \sqrt{2 - \sec^{-1}x}$ to be defined,the expression inside the square root must be non-negative:$2 - \sec^{-1}x \ge 0$$\s

Vedclass · Accepted Answer

$\left( -\infty, \sec 2 \right] \cup \left[ 1, \infty \right)$

Vedclass · Answer

(C) For the function $f(x) = \sqrt{2 - \sec^{-1}x}$ to be defined,the expression inside the square root must be non-negative:$2 - \sec^{-1}x \ge 0$$\sec^{-1}x \le 2$Also,the domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$.Since $\sec^{-1}x \le 2$,and the range of $\sec^{-1}x$ is $[0, \pi] \setminus \{\pi/2\}$,we consider the inequality $\sec^{-1}x \le 2$.For $x \ge 1$,$\sec^{-1}x \in [0, \pi/2)$,so $\sec^{-1}x \le 2$ is always true for all $x \in [1, \infty)$.For $x \le -1$,$\sec^{-1}x \in (\pi/2, \pi]$. We need $\sec^{-1}x \le 2$. Since $2 Taking the secant of both sides (noting that $\sec$ is increasing on $[0, \pi/2)$ and decreasing on $(\pi/2, \pi]$ is not directly applicable,but we know $\sec^{-1}x \le 2 \implies x \le \sec 2$ for $x \le -1$ because $\sec 2$ is negative).Thus,the domain is $(-\infty, \sec 2] \cup [1, \infty)$.

The domain of the function $f(x) = \sqrt{2 - \sec^{-1}x}$ is:

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