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If the function $f(x)=x^3+e^{\frac{x}{2}}$ and $g(x)=f^{-1}(x)$, then the value of $g^{\prime}(1)$ is
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Verified Answer
The correct answer is:
2
Given, $g\{f(x)\}=x$
$$
\begin{array}{lc}
\Rightarrow & g^{\prime}\{f(x)\} f^{\prime}(x)=1 \\
\text { If } & f(x)=1 \Rightarrow x=0, f(0)=1
\end{array}
$$
Substitute $x=0$ in Eq. (i), we get
$$
\begin{aligned}
& g^{\prime}(1)=\frac{1}{f^{\prime}(0)} \\
& \Rightarrow \quad g^{\prime}(1)=2 \\
& {\left[\because f^{\prime}(x)=3 x^2+\frac{1}{2} e^{x / 2} \Rightarrow f^{\prime}(0)=\frac{1}{2}\right]} \\
&
\end{aligned}
$$
$$
\begin{array}{lc}
\Rightarrow & g^{\prime}\{f(x)\} f^{\prime}(x)=1 \\
\text { If } & f(x)=1 \Rightarrow x=0, f(0)=1
\end{array}
$$
Substitute $x=0$ in Eq. (i), we get
$$
\begin{aligned}
& g^{\prime}(1)=\frac{1}{f^{\prime}(0)} \\
& \Rightarrow \quad g^{\prime}(1)=2 \\
& {\left[\because f^{\prime}(x)=3 x^2+\frac{1}{2} e^{x / 2} \Rightarrow f^{\prime}(0)=\frac{1}{2}\right]} \\
&
\end{aligned}
$$
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