SOLUTION(a) domain: \(x > 0\)(b) find the coordinates of the minimum point on the graph of \(g\)\(g"\left( x \right) = x \cdot \frac{1}{x} + \ln x = 1 + \ln x = 0\;\;\; \Rightarrow \;\;\;\ln x = - 1\;\;\; \Rightarrow \;\;\;x = {{\rm{e}}^{ - 1}} = \frac{1}{{\rm{e}}}\)\(g\left( {\frac{1}{{\rm{e}}}} \right) = \frac{1}{{\rm{e}}}\ln \left( {\frac{1}{{\rm{e}}}} \right) = \frac{1}{{\rm{e}}}\left( { - 1} \right) = - \frac{1}{{\rm{e}}}\)thus, coordinates of minimum point are \(\left( {\frac{1}{{\rm{e}}}, - \frac{1}{{\rm{e}}}} \right)\)therefore, the range of \(g\) is \(y \ge - \frac{1}{{\rm{e}}}\)(c) A unique line exists that passes through \(\left( {0, - \,c} \right)\), where \(c > 0\), and is tangent to the graph of g. Let P be the point of tangency. If the x-coordinate of P is p, then the y-coordinate of P is \(p\ln p\). Given that...