Problem of the Week #3 (2 Dec 2019)

SOLUTION

(a)   domain: \(x > 0\)

(b)   find the coordinates of the minimum point on the graph of \(g\)

\(g'\left( x \right) = x \cdot \frac{1}{x} + \ln x = 1 + \ln x = 0\;\;\; \Rightarrow \;\;\;\ln x = - 1\;\;\; \Rightarrow \;\;\;x = {{\textrm{e}}^{ - 1}} = \frac{1}{{\textrm{e}}}\)

\(g\left( {\frac{1}{{\textrm{e}}}} \right) = \frac{1}{{\textrm{e}}}\ln \left( {\frac{1}{{\textrm{e}}}} \right) = \frac{1}{{\textrm{e}}}\left( { - 1} \right) = - \frac{1}{{\textrm{e}}}\)

thus, coordinates of minimum point are \(\left( {\frac{1}{{\textrm{e}}}, - \frac{1}{{\textrm{e}}}} \right)\)

therefore, the range of \(g\) is \(y \ge - \frac{1}{{\textrm{e}}}\)

(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 \(g'\left( x \right) = 1 + \ln x\) then the gradient of the tangent at P is \(g'\left( p \right) = 1 + \ln p\). Since the tangent passes through \(\left( {p,p\ln p} \right)\) and \(\left( {0, - \,c} \right)\) then the gradient of the tangent can also be computed by \(\frac{{p\ln p + c}}{{p - 0}}\). Thus, \(\frac{{p\ln p + c}}{p} = 1 + \ln p\;\;\; \Rightarrow \;\;\;p\ln p + c = p + p\ln p\;\;\; \Rightarrow \;\;\;p = c\).  Hence, the point of tangency, in terms of c, is \(\left( {c,c\ln c} \right)\) and the gradient of the tangent is \(g'\left( c \right) = 1 + \ln c\).  Using the point-gradient form, the equation of the tangent line is: \(y - \left( { - \,c} \right) = \left( {1 + \ln c} \right)\left( {x - 0} \right)\;\;\; \Rightarrow \;\;\;y = x\left( {1 + \ln c} \right) - c\) ;  or \({y = x - c + x\ln c}\)

Although a GDC is not allowed for the solution to this problem, below is a video showing a dynamic graph on a TI-Nspire confirming the answer found in part (c) above. The graphs of \(g\left( x \right) = x\ln x\) and the tangent line \(y = x\left( {1 + \ln c} \right) - c\), with a slider for \(c\). Note how the x-coordinate of the point of tangency is equal to the value of \(c\), and the y-intercept of the tangent line is equal to the opposite of \(c\).