If $\mathrm{f}(x)=x^2+1$ and $\mathrm{g}(x)=\frac{1}{x}$, then the value of $\mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(x))))$ at $x=1$ is

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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{f}(x)=x^2+1, & \mathrm{~g}(x)=\frac{1}{x} \ 	herefore \quad \mathrm{f}(\mathrm{g}(\mathrm{g}(\mathrm{f}(x)))) & =\mathrm{f}\left(\mathrm{g}\left(\mathrm{g}\left(x^2+1ight)ight)ight) \ & =\mathrm{f}\left(\mathrm{g}\left(\frac{1}{x^2+1}ight)ight) \ & =\mathrm{f}\left(x^2+1ight) \ & =\left(x^2+1ight)^2+1 \end{aligned}$ $	herefore \quad$ At $x=1$, we get the value of above function $\begin{aligned} & =\left[(1)^2+1ight]^2+1 \ & =5 \end{aligned}$

If $\mathrm{f}(x)=x^2+1$ and $\mathrm{g}(x)=\frac{1}{x}$, then the value of…

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