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Question Number 135838 by abdurehime last updated on 16/Mar/21

Answered by liberty last updated on 16/Mar/21

$$f\left(x\right)=\frac{x}{x^2+k}\Rightarrow \ln y=\ln x-\ln (x^2+k)$$$$\frac{y^{\prime }}{y}=\frac{1}{x}-\frac{2x}{x^2+k}$$$$\frac{y^{\prime }}{y}=\frac{k-x^2}{x(x^2+k)};y^{\prime }=\frac{x}{(x^2+k)}.\left(\frac{k-x^2}{x(x^2+k)}\right)$$$$y^{\prime }=\frac{k-x^2}{{(x^2+k)}^2}=0$$$$wegetcriticalpointx=\pm \sqrt{k}$$$$\begin{array}{|cc|}\hline interval& sign\\ x<-\sqrt{k}& -\\ -\sqrt{k}<x<\sqrt{k}& +\\ x>\sqrt{k}& -\\ \hline\end{array}$$$$wegetmaximumvalue$$$$whenx=\sqrt{k}=2\Rightarrow k=4$$

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