Find point of inflection at y=X⁴−2X²

$$y = x^4 - 2x^2$$ $$y' = \frac{d}{dx}(x^4 - 2x^2) = 4x^3 - 4x$$

$$12x^2 - 4 = 0$$ $$12x^2 = 4$$ $$x^2 = \frac{4}{12}$$ $$x^2 = \frac{1}{3}$$ $$x = \pm \sqrt{\frac{1}{3}}$$ $$x = \pm \frac{1}{\sqrt{3}}$$

Rationalize the denominator:

$$x = \pm \frac{\sqrt{3}}{3}$$

We need to check the sign of $y''$ in intervals around $x = -\frac{\sqrt{3}}{3}$ and $x = \frac{\sqrt{3}}{3}$.

• For $x < -\frac{\sqrt{3}}{3}$ (e.g., $x=-1$): $y''(-1) = 12(-1)^2 - 4 = 12 - 4 = 8 > 0$. (Concave up)
• For $-\frac{\sqrt{3}}{3} < x < \frac{\sqrt{3}}{3}$ (e.g., $x=0$): $y''(0) = 12(0)^2 - 4 = -4 < 0$. (Concave down)
• For $x > \frac{\sqrt{3}}{3}$ (e.g., $x=1$): $y''(1) = 12(1)^2 - 4 = 12 - 4 = 8 > 0$. (Concave up) Since the concavity changes at both $x = -\frac{\sqrt{3}}{3}$ and $x = \frac{\sqrt{3}}{3}$, these are indeed the $x$-coordinates of the points of inflection.

For $x = \frac{\sqrt{3}}{3}$:

$$y = \left(\frac{\sqrt{3}}{3}\right)^4 - 2\left(\frac{\sqrt{3}}{3}\right)^2$$ $$y = \left(\frac{3}{9}\right)^2 - 2\left(\frac{3}{9}\right)$$ $$y = \left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right)$$ $$y = \frac{1}{9} - \frac{2}{3}$$ $$y = \frac{1}{9} - \frac{6}{9}$$ $$y = -\frac{5}{9}$$

So, one point of inflection is $\left(\frac{\sqrt{3}}{3}, -\frac{5}{9}\right)$.

For $x = -\frac{\sqrt{3}}{3}$:
Since the powers in the original function ($x^4$ and $x^2$) are even, $y$ will be the same for $x = -\frac{\sqrt{3}}{3}$ as for $x = \frac{\sqrt{3}}{3}$.

$$y = \left(-\frac{\sqrt{3}}{3}\right)^4 - 2\left(-\frac{\sqrt{3}}{3}\right)^2 = \frac{1}{9} - \frac{6}{9} = -\frac{5}{9}$$

So, the other point of inflection is $\left(-\frac{\sqrt{3}}{3}, -\frac{5}{9}\right)$.

The points of inflection are $\boxed{\left(\frac{\sqrt{3}}{3}, -\frac{5}{9}\right) \text{ and } \left(-\frac{\sqrt{3}}{3}, -\frac{5}{9}\right)}$.