fun-lim

Question: Suppose that $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2\sqrt{x}+x$, then $f(x)$ is



1) $1+2{{x}^{2}}$
2) $2+{{x}^{2}}$
3) $1+x$
4) $2+x$
Solution: Explanation: No Explanation
Functions

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