InterviewSolution
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InterviewSolution
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consider the function f(x) =1+(1)/(x)^(x) The range of the function f(x) is |
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Answer» `(0,poo)` f(x) is defined if`1+1/xgt0 or (x+1)/(x)gt0` or `(-oo,-1)cup(0,oo)` Now f(X)=`(1+(1)/(x))^(x)ln (1+(1)/(x))+(x)/(1+(1)/(x))(-1)/(x^(2))` `=(1+(1)/(x))^(x)ln(1+(1)/(x))-(1)/(x+1)` Now `(1+(1)/(x))^(x)` is always positive Hence the sign of f(X) Depends on the sign of ln `(1+(1)/(x))-(1)/(1+x)` Let `g(x)=In (1+(1)/(x))-(1)/(x+1)` `g(x)=(1)/(1+(1))(x)(-1)/(x^(2))+(1)/(x+1)^(2)=(-1)/(x(x+1)^(2)` (i)for `x in (0,oo)g(x)lt0` Thus g(X) is monotonically decreasing for x in `(0,oo)` or `g(X) gt underset(xrarroo)limg(x)` or `g(X)gt0, so f(X) gt0` (ii)for `x in (-oo,-1),g(x)gt0` Thus g(X) is monotonically INCREASING for `x in (-oo,-1)` or `g(X) gt underset(xrarroo)limg(X)gt0` `THEREFORE f(x)gt0` Hence FORM (i) nd (ii) we get `f(X) gt0 FORALL x in (-oo,-1)cup(0,oo)` Thus f(x) is montonically increasing in its domain Also `underset(xrarroo)lim(1+(1)/(x))^(x)`=e `underset(xrarr0)lim(1+(1)/(x))^(x)=1 and underset(xrarr-1)lim(1+(1)/(x))^(x)=oo` The graph of f(X) is shown in figure  Range is `y in (1,oo)-{e}` |
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