If f(x) = x^{11} + sin^3(35x) + 111x,then the | vedclass

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(D) Given $f(x) = x^{11} + \sin^3(35x) + 111x$. Since $x^{11}$,$\sin^3(35x)$,and $111x$ are all odd functions,their sum $f(x)$ is an odd function,i.e.

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$f(0)$

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(D) Given $f(x) = x^{11} + \sin^3(35x) + 111x$. Since $x^{11}$,$\sin^3(35x)$,and $111x$ are all odd functions,their sum $f(x)$ is an odd function,i.e.,$f(-x) = -f(x)$. If $f(x)$ is an odd function,then its inverse $f^{-1}(x)$ is also an odd function,i.e.,$f^{-1}(-y) = -f^{-1}(y)$. We need to calculate $S = f^{-1}(\sin \frac{\pi}{5}) + f^{-1}(\sin \frac{6\pi}{5}) + f^{-1}(\sin \frac{\pi}{7}) + f^{-1}(\sin \frac{8\pi}{7})$. Note that $\sin \frac{6\pi}{5} = \sin(\pi + \frac{\pi}{5}) = -\sin \frac{\pi}{5}$. Similarly,$\sin \frac{8\pi}{7} = \sin(\pi + \frac{\pi}{7}) = -\sin \frac{\pi}{7}$. Substituting these into the expression: $S = f^{-1}(\sin \frac{\pi}{5}) + f^{-1}(-\sin \frac{\pi}{5}) + f^{-1}(\sin \frac{\pi}{7}) + f^{-1}(-\sin \frac{\pi}{7})$. Using the property $f^{-1}(-y) = -f^{-1}(y)$: $S = f^{-1}(\sin \frac{\pi}{5}) - f^{-1}(\sin \frac{\pi}{5}) + f^{-1}(\sin \frac{\pi}{7}) - f^{-1}(\sin \frac{\pi}{7}) = 0$. Since $f(0) = 0^{11} + \sin^3(0) + 111(0) = 0$,the result $0$ is equal to $f(0)$.

If $f(x) = x^{11} + \sin^3(35x) + 111x$,then the value of $f^{-1}(\sin \frac{\pi}{5}) + f^{-1}(\sin \frac{6\pi}{5}) + f^{-1}(\sin \frac{\pi}{7}) + f^{-1}(\sin \frac{8\pi}{7})$ is equal to

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