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  4. Let f: R arrow R be defined as f ( x )=(2 x -3 π)3+(4/3) x + cos x and g = f -1, then

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Q. Let $f : R \rightarrow R$ be defined as $f ( x )=(2 x -3 \pi)^3+\frac{4}{3} x +\cos x$ and $g = f ^{-1}$, then

Continuity and Differentiability

Solution:

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Now, $g^{\prime}(2 \pi)=\frac{1}{f^{\prime}\left(\frac{3 \pi}{2}\right)}=\frac{1}{\frac{7}{3}}=\frac{3}{7}$
$f^{\prime}(x)=2 \times 3(2 x-3 \pi)^2+\frac{4}{3}-\sin x $
$\therefore f^{\prime}\left(\frac{3 \pi}{2}\right)=\frac{4}{3}+1=\frac{7}{3}$
Also, $g^{\prime \prime}(2 \pi)=\frac{-f^{\prime \prime}\left(\frac{3 \pi}{2}\right)}{\left(f^{\prime}\left(\frac{3 \pi}{2}\right)\right)^3}=0$
$ f ^{\prime \prime}( x )=12 \times 2(2 x -3 \pi)+0-\cos x $
$\therefore f ^{\prime \prime}\left(\frac{3 \pi}{2}\right)=0$