If f(x) = sqrt{2x - 1} + 5 cos^{-1}left(frac{ | vedclass

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(B) For the function $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$ to be defined,both terms must be defined.For $\sqrt{2x - 1}$ to

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$\left[\frac{1}{2}, 2\right]$

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(B) For the function $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$ to be defined,both terms must be defined.For $\sqrt{2x - 1}$ to be defined,we must have $2x - 1 \geq 0$,which implies $x \geq \frac{1}{2}$.For $\cos^{-1}\left(\frac{2x - 1}{3}\right)$ to be defined,the argument must satisfy $-1 \leq \frac{2x - 1}{3} \leq 1$.Multiplying by $3$,we get $-3 \leq 2x - 1 \leq 3$.Adding $1$ to all parts,we get $-2 \leq 2x \leq 4$.Dividing by $2$,we get $-1 \leq x \leq 2$.The domain is the intersection of $x \geq \frac{1}{2}$ and $-1 \leq x \leq 2$.Thus,the domain is $\left[\frac{1}{2}, 2\right]$.

If $f(x) = \sqrt{2x - 1} + 5 \cos^{-1}\left(\frac{2x - 1}{3}\right)$,then the domain of the function $f(x)$ is

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