If the real valued function f(x)=sin ^{-1}(x^ | vedclass

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(D) The function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is defined when both terms are defined.For $\sin ^{-1}(x^2-1)$, we require $-1 \leq x^2-1 \

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$4$

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(D) The function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is defined when both terms are defined.For $\sin ^{-1}(x^2-1)$, we require $-1 \leq x^2-1 \leq 1$, which implies $0 \leq x^2 \leq 2$, so $x \in [-\sqrt{2}, \sqrt{2}]$.For $\log _3(3^x-2)$, we require $3^x-2 > 0$, which implies $3^x > 2$, so $x > \log _3 2$.The domain of $f(x)$ is the intersection: $[-\sqrt{2}, \sqrt{2}] \cap (\log _3 2, \infty) = (\log _3 2, \sqrt{2}]$.The function is not defined for $x \in (-\infty, \log _3 2] \cup (\sqrt{2}, \infty)$.Comparing this with $(-\infty, a] \cup (b, \infty)$, we get $a = \log _3 2$ and $b = \sqrt{2}$.Thus, $3^a + b^2 = 3^{\log _3 2} + (\sqrt{2})^2 = 2 + 2 = 4$.

Function	Range
$A. f(x) = \|x\|$	$I. [0, \infty)$
$B. f(x) = x^2$	$II. \mathbb{R}$
$C. f(x) = x^3$	$III. [0, \infty)$
$D. f(x) = \text{sgn}(x)$	$IV. \{-1, 0, 1\}$

If the real valued function $f(x)=\sin ^{-1}(x^2-1)-3 \log _3(3^x-2)$ is not defined for all $x \in(-\infty, a] \cup(b, \infty)$, then $3^a+b^2=$

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