Let f(x)=x+frac{1}{x} and g(x)=e^{x}. The f(g(0))=

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Step:1 ：Let (f(x)=x+frac{1}{x}) and (g(x)=e^{x}). We want to find the value of (f(g(0))).Step:2 ：First, we need to find the value of (g(0)). The function (g(x)) is an exponential function with base (e). When (x=0), (e^{x}) equals 1. So, (g(0)=1).Step:3 ：Then, we substitute (g(0)) into the function (f(x)), which is (f(x)=x+frac{1}{x}). So, (f(g(0))=f(1)=1+frac{1}{1}=2).Step:4 ：Final Answer: (f(g(0))=boxed{2})

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Step: 1 ：Let (f(x)=x+frac{1}{x}) and (g(x)=e^{x}). We want to find the value of (f(g(0))).Step: 2 ：First, we need to find the value of (g(0)). The function (g(x)) is an exponential function with base (e). When (x=0), (e^{x}) equals 1. So, (g(0)=1).Step: 3 ：Then, we substitute (g(0)) into the function (f(x)), which is (f(x)=x+frac{1}{x}). So, (f(g(0))=f(1)=1+frac{1}{1}=2).Step: 4 ：Final Answer: (f(g(0))=boxed{2})

Let $f(x)=x+\frac{1}{x}$ and $g(x)=e^{x}$. The $f(g(0))=$

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