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Draw the graph of f(x)="ln" (1-"ln "x). Find thepoint of inflection. |
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Answer» Solution :`f(X) = "ln"(1-"ln "x)` `f(x)` is defined if `1-"ln "x gt0` or `"ln " x lt 1` or `0 lt x lt e` So the DOMAIN is (0,e). ALSO, `f^(')(x)=-1/((1-"ln "x). 1/x lt 0` So `f(x)` is continous and decreasing `AA x in (0,e)`. Further `underset(xto0^(+))"lim""ln "(1-" ln "x)="ln "(1+infty)=infty` `underset(x to e^(-))"lim" "ln "(1-"ln"x) = "ln "(1-1^(-))="ln "0^(+)=-infty` For the point of inflection, `f^('')(x) = (-"ln "x)/(x^(2)(1-"ln "x)^(2))` Clearly, `f^('')(1)=0`, so `x=1` is a point of inflection. From the above discussion, the graph of `y=f(x)` is as shown in the following figure. From the graph, `y=0` and y=e are asymptotoes. |
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