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by **stuart clark** on Thu Mar 03, 2011 3:23 pm

If $f(x) = (2x-\pi)^3+2x-cosx$ .Then Calculate $\displaystyle\frac{d}{dx}\left(f^{-1}(x)\right)$ at $x=\pi$

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by **nona.m.nona** on Thu Mar 03, 2011 4:15 pm

stuart clark wrote:If $f(x) = (2x-\pi)^3+2x-cosx$ .Then Calculate $\displaystyle\frac{d}{dx}\left(f^{-1}(x)\right)$ at $x=\pi$

The first step will be to find the value $a$ for which $f(a)\, =\, \pi$ , since then $f^{-1}(\pi)\,=\,a$ . Hint: Consider values for which the cosine will be zero.

Then use the fact that $\left[f^{-1}(\pi)\right] = \frac{1}{f'(f^{-1}(\pi))}$ , plugging in the value you have found for $a\, =\, f^{-1}(\pi)$ .

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