Find the inflection points of the curve y = (x^4)/(x³-1)

Find the inflection points of the curve $y = \dfrac{x^{4}}{x^{3}-1}$, $x\neq1$.

Step 1: The first derivative

\[ y' = \frac{4x^{3}\bigl(x^{3}-1\bigr)-x^{4}\cdot3x^{2}}{\bigl(x^{3}-1\bigr)^{2}} = \frac{x^{6}-4x^{3}}{\bigl(x^{3}-1\bigr)^{2}} = \frac{x^{3}\bigl(x^{3}-4\bigr)}{\bigl(x^{3}-1\bigr)^{2}}. \]

Step 2: The second derivative

With $P = x^{6}-4x^{3}$ and $Q = \bigl(x^{3}-1\bigr)^{2}$, so $P' = 6x^{2}\bigl(x^{3}-2\bigr)$ and $Q' = 6x^{2}\bigl(x^{3}-1\bigr)$, the common factor $6x^{2}\bigl(x^{3}-1\bigr)$ cancels one power of the denominator:

\[ y'' = \frac{6x^{2}\left[\bigl(x^{3}-2\bigr)\bigl(x^{3}-1\bigr)-x^{3}\bigl(x^{3}-4\bigr)\right]}{\bigl(x^{3}-1\bigr)^{3}}, \] and the bracket collapses: \[\bigl(x^{6}-3x^{3}+2\bigr)-\bigl(x^{6}-4x^{3}\bigr) = x^{3}+2, \qquad\text{so}\qquad y'' = \frac{6x^{2}\bigl(x^{3}+2\bigr)}{\bigl(x^{3}-1\bigr)^{3}}. \tag{1}\]

Step 3: Candidates

\[ y'' = 0 \;\Longleftrightarrow\; x = 0 \ \text{ (a double root) } \ \text{ or } \ x = -\sqrt[3]{2}. \]

Step 4: Sign of $y''$

\[ \begin{array}{c|c|c|c} \text{interval} & x^{3}+2 & \bigl(x^{3}-1\bigr)^{3} & y'' \\ \hline x\lt -\sqrt[3]{2} & - & - & + \\ -\sqrt[3]{2}\lt x\lt 0 & + & - & - \\ 0\lt x\lt 1 & + & - & - \\ x\gt 1 & + & + & + \end{array} \]

At $x = 0$ the factor $x^{2}$ touches zero without changing sign, so $y''$ keeps its sign: not an inflection. At $x = -\sqrt[3]{2}$ the sign changes from $+$ to $-$: an inflection. (The point $x = 1$ is a pole, not a point of the curve.)

Step 5: The coordinate

With $x_{0} = -\sqrt[3]{2}$ we have $x_{0}^{3} = -2$ and $x_{0}^{4} = 2^{4/3} = 2\sqrt[3]{2}$, so

\[ y\bigl(x_{0}\bigr) = \frac{2\sqrt[3]{2}}{-2-1} = -\frac{2\sqrt[3]{2}}{3}. \]

Step 6: Verification

\[y''(-2) = \frac{6\cdot4\cdot(-6)}{(-9)^{3}} = 0.1975\gt 0; \qquad y''(-1) = \frac{6\cdot1\cdot1}{(-2)^{3}} = -0.75\lt 0 \;\;\checkmark\]

\[y''(0.5) = \frac{6\cdot0.25\cdot2.125}{(-0.875)^{3}} = -4.758\lt 0:\ \text{the same sign as just left of } 0 \;\;\checkmark\]

\[y''(2) = \frac{6\cdot4\cdot10}{7^{3}} = 0.6997\gt 0 \;\;\checkmark\]