Let $f(x) = x + \sin x$. Suppose $g$ denotes the inverse function of $f$. If the value of $g'\left(\frac{\pi}{4} + \frac{1}{\sqrt{2}}\right)$ is $l$ then $2l = $
Step-by-Step Solution
Key Concept: The derivative of an inverse function is the reciprocal of the original function's derivative at the corresponding point.
Given $f(x) = y = x + \sin x$, we have $\frac{dy}{dx} = 1 + \cos x$ and $g'(y) = \frac{dx}{dy} = \frac{1}{1+\cos x}$. At the point where $y = \frac{\pi}{4} + \frac{1}{\sqrt{2}}$, we get $x = \frac{\pi}{4}$, so $g'\left(\frac{\pi}{4} + \frac{1}{\sqrt{2}}\right) = \frac{1}{1+\cos\frac{\pi}{4}} = \frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = 2 - \sqrt{2}$.
Correct Answer: 4