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Q.
Let $f: R \rightarrow R$ defined as $f(x)=x^5+e^{\frac{x}{2}}+e^{x^3}$ and $g(x)=f^{-1}(x)$, then the value of $g^{\prime}(2)$, is
Continuity and Differentiability
Solution:
$ fC ( x )= x ^5+ e ^{ x / 2}+ e ^{ x ^3}$
$f^{\prime}(x)=5 x^4+\frac{1}{2} e^{x / 2}+3 x^2 e^{x^3}>0$
$\therefore f ( x )$ is one one funciton Clearly given function will be onto function as its range is $R$.
$\because g(f(x))=x \Rightarrow g^{\prime}(f(x)) \cdot f^{\prime}(x)=1 \Rightarrow g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} $
$f(x)=2 \text { when } x=0$
$\therefore g^{\prime}(2)=\frac{1}{f^{\prime}(0)}=\frac{1}{\left(\frac{1}{2}\right)}=2 $