Let f be a real valued function defined by f( | vedclass

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(A) For the function $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$ to be defined,the arguments of $

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$[-4, 4]$

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(A) For the function $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$ to be defined,the arguments of $\sin^{-1}$ and $\cos^{-1}$ must lie in the interval $[-1, 1]$.$1$. For $\sin^{-1} \left( \frac{1 - |x|}{3} \right)$,we have:$-1 \le \frac{1 - |x|}{3} \le 1$$-3 \le 1 - |x| \le 3$$-4 \le -|x| \le 2$$-2 \le |x| \le 4$Since $|x| \ge 0$,this implies $0 \le |x| \le 4$.$2$. For $\cos^{-1} \left( \frac{|x| - 3}{5} \right)$,we have:$-1 \le \frac{|x| - 3}{5} \le 1$$-5 \le |x| - 3 \le 5$$-2 \le |x| \le 8$Since $|x| \ge 0$,this implies $0 \le |x| \le 8$.Taking the intersection of both conditions:$0 \le |x| \le 4$This inequality is equivalent to $-4 \le x \le 4$.Thus,the domain of $f(x)$ is $[-4, 4]$.

Let $f$ be a real valued function defined by $f(x) = \sin^{-1} \left( \frac{1 - |x|}{3} \right) + \cos^{-1} \left( \frac{|x| - 3}{5} \right)$. Then the domain of $f(x)$ is given by:

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