Given that $f(x) = 2x + \cos(x)$ is one-to-one, use the formula $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ to find $(f^{-1})'(1)$. $(f^{-1})'(1) = $
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To find the inverse function, we need to solve the equation y = f(x) for x. In this case, we have y = 2 + cos(x). To solve for x, we subtract 2 from both sides and take the inverse cosine of both sides: y - 2 = cos(x) cos^(-1)(y - 2) = x So the inverse function is Show more…
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