If f(x) = (2x - 3pi)^5 + frac{4}{3}x + cos x | vedclass

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(B) Given $f(x) = (2x - 3\pi)^5 + \frac{4}{3}x + \cos x$. Since $g$ is the inverse of $f$,we have $g(f(x)) = x$. Differentiating both sides with respe

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$\frac{3}{7}$

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(B) Given $f(x) = (2x - 3\pi)^5 + \frac{4}{3}x + \cos x$. Since $g$ is the inverse of $f$,we have $g(f(x)) = x$. Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$,which implies $g'(f(x)) = \frac{1}{f'(x)}$. We need to find $g'(2\pi)$. Let $f(x) = 2\pi$. $(2x - 3\pi)^5 + \frac{4}{3}x + \cos x = 2\pi$. By observation,if $x = \frac{3\pi}{2}$,then $f(\frac{3\pi}{2}) = (2(\frac{3\pi}{2}) - 3\pi)^5 + \frac{4}{3}(\frac{3\pi}{2}) + \cos(\frac{3\pi}{2}) = (0)^5 + 2\pi + 0 = 2\pi$. Now,find $f'(x)$: $f'(x) = 5(2x - 3\pi)^4 \cdot 2 + \frac{4}{3} - \sin x = 10(2x - 3\pi)^4 + \frac{4}{3} - \sin x$. At $x = \frac{3\pi}{2}$,$f'(\frac{3\pi}{2}) = 10(0)^4 + \frac{4}{3} - \sin(\frac{3\pi}{2}) = 0 + \frac{4}{3} - (-1) = \frac{4}{3} + 1 = \frac{7}{3}$. Therefore,$g'(2\pi) = \frac{1}{f'(\frac{3\pi}{2})} = \frac{1}{7/3} = \frac{3}{7}$.

If $f(x) = (2x - 3\pi)^5 + \frac{4}{3}x + \cos x$ and $g$ is the inverse of $f$,then $g'(2\pi) = ?$

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