The domain of the function f(x) = sec^{-1}(3x | vedclass

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(D) The function is $f(x) = \sec^{-1}(3x - 4) + 	anh^{-1}\left(\frac{x + 3}{5}ight)$.For $\sec^{-1}(3x - 4)$ to be defined,we must have $|3x - 4| \

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$(-8, 1] \cup \left[\frac{5}{3}, 2\right)$

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(D) The function is $f(x) = \sec^{-1}(3x - 4) + 	anh^{-1}\left(\frac{x + 3}{5}ight)$.For $\sec^{-1}(3x - 4)$ to be defined,we must have $|3x - 4| \geq 1$.This implies $3x - 4 \leq -1$ or $3x - 4 \geq 1$.$3x \leq 3 \Rightarrow x \leq 1$ and $3x \geq 5 \Rightarrow x \geq \frac{5}{3}$.So,the domain of the first part is $x \in (-\infty, 1] \cup [\frac{5}{3}, \infty)$.For $	anh^{-1}\left(\frac{x + 3}{5}ight)$ to be defined,we must have $-1 $-5 $-8 Taking the intersection of $x \in (-\infty, 1] \cup [\frac{5}{3}, \infty)$ and $x \in (-8, 2)$:$x \in (-8, 1] \cup [\frac{5}{3}, 2)$.

The domain of the function $f(x) = \sec^{-1}(3x - 4) + \tanh^{-1}\left(\frac{x + 3}{5}\right)$ is

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