se f(x) = ln (1 + e2x) encontre f(0)

Considerando \(f(x)=\ln(1+e^{2x} )\), o valor de \(f(0)\) é:

\(\Longrightarrow f(0)=\ln(1+e^{2\cdot 0} )\)

\(\Longrightarrow f(0)=\ln(1+e^{0} )\)

\(\Longrightarrow f(0)=\ln(1+1 )\)

\(\Longrightarrow \fbox {$ f(0)=\ln(2) \approx 0,693 $}\)