# P.o.t.W. #3 Solution

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...

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