Question

\(f(x)=\frac{x}{1-x^{2}}\) $f^{\prime}(x)=\frac{\left(1-x^{2}\right)-x(-2 x)}{\left(1-x^{2}\right)^{2}}=\frac{1+x^{2}}{\left(1-x^{2}\right)^{2}}=1$ \(1+x^{2}=1+x^{4}-2 x^{2}\) \(\Rightarrow \quad x^{2}\left(x^{2}-3\right)=0\) \(\Rightarrow x=0, \pm \sqrt{3}\) $P t \rightarrow(0,0),\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right)\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)$

Short answer

Solution: The tangent line is horizontal at the points: $$(0,0),\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right),\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)$$

Step-by-step solution

Find the derivative \(f^{\prime}(x)\).

To find the derivative of the function, we need to use the quotient rule: $$(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$$ Here, \(u=x\) and \(v=1-x^2\). Thus, we have $$u' = \frac{d(x)}{dx} = 1$$ $$v' = \frac{d(1-x^{2})}{dx}=-2x$$ Now we can apply the quotient rule to find the derivative: $$f^{\prime}(x) = \frac{u'v - uv'}{v^2} = \frac{(1)(1-x^2) - x(-2x)}{(1-x^2)^2}$$

Set \(f^{\prime}(x)=1\) and solve for \(x\).

As mentioned in the analysis, we want to find the points where the tangent is horizontal, i.e., \(f^{\prime}(x)=1\). So, we set the derivative equal to \(1\) and solve for \(x\): $$\frac{1+x^2}{(1-x^2)^2} = 1$$ $$1+x^2=1+x^4-2x^2$$ $$\Rightarrow x^2(x^2-3)=0$$ $$\Rightarrow x=0, \pm\sqrt{3}$$

Calculate the corresponding \(y\)-values.

Now, we need to find the corresponding \(y\)-values for each value of \(x\). We plug each \(x\) value back into the original function: $$f(0) = \frac{0}{1 - 0^2} = 0$$ $$f(\sqrt{3}) = \frac{\sqrt{3}}{1 - (\sqrt{3})^2} = \frac{\sqrt{3}}{1 - 3} = \frac{\sqrt{3}}{2}$$ $$f(-\sqrt{3}) = \frac{-\sqrt{3}}{1 - (-\sqrt{3})^2} = \frac{-\sqrt{3}}{1 - 3} = \frac{\sqrt{3}}{2}$$

Write down the points in the form of a tuple \((x,y)\).

Now we write down the points on the curve with horizontal tangent lines: $$(0,0),\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right),\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)$$