Mathematics - Application of Derivative Question with Solution | TestHub

$\left(x^2+1\right)(y-3)=x\Rightarrow y=3+\frac{x}{x^2+1}$ $\begin{array}{rlr}& m=\frac{dy}{dx}=\frac{1-x^2}{{\left(1+x^2\right)}^2}& \frac{dm}{dx}=\frac{-2x{\left(1+x^2\right)}^2-\left(1-x^2\right)\cdot 2\left(1+x^2\right)\cdot 2x}{{\left(1+x^2\right)}^4}=\frac{2x\left(x^2-3\right)}{{\left(1+x^2\right)}^3}\end{array}$ For extremum $\frac{dm}{dx}=0\Rightarrow x=0,x=\pm \sqrt{3}$ $\frac{dm}{dx}$ changes from positive to negative at $x=0$, so $m$ is maximum at $x=0$, where $m=1$. (At $x=\pm \sqrt{3},m=-\frac{1}{8}$, the minimum value.) $x=0\Rightarrow y=3\therefore$ the point is $(0,3)$