Find the set of points at which the function y = (x³+

Find the set of points at which the function $\displaystyle y = \frac{x^{3}+\alpha}{x^{2}-\alpha}$, $\alpha\gt 0$, attains extreme values.

Step 1: The derivative

The function is defined for $x\neq\pm\sqrt{\alpha}$, and

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

Step 2: Two kinds of critical point

Either $x = 0$, or $x^{3}-3\alpha x-2\alpha = 0$. The cubic does not factor over the rationals, but it need not be solved: it is linear in $\alpha$, so it can be read as a formula for the parameter,

\[ x^{3} = \alpha\bigl(3x+2\bigr) \;\Longleftrightarrow\; \alpha = \frac{x^{3}}{3x+2}. \tag{2} \]

The requirement $\alpha\gt 0$ restricts $x$ to $x\gt 0$ or $x\lt -\tfrac23$.

Step 3: Eliminate $\alpha$

Substituting (2) into the function itself: \[y = \frac{x^{3}+\dfrac{x^{3}}{3x+2}}{x^{2}-\dfrac{x^{3}}{3x+2}} = \frac{x^{3}\bigl(3x+3\bigr)}{x^{2}\bigl(2x+2\bigr)} = \frac{3x\bigl(x+1\bigr)}{2\bigl(x+1\bigr)} = \frac{3x}{2} \qquad (x\neq-1). \tag{3}\]

So every critical point of this second kind lies on the straight line $y = \dfrac32x$. The exclusion $x\neq-1$ is genuine: $x = -1$ forces $\alpha = 1$, and then $x^{2}-\alpha = 0$ — the point is a pole, not a point of the graph.

Step 4: The critical point $x = 0$

Here $y(0) = \dfrac{\alpha}{-\alpha} = -1$ for every $\alpha\gt 0$, so all these points coincide at $\bigl(0,-1\bigr)$, which is not on the line. Near the origin (1) gives

\[ y' \approx \frac{x\bigl(-2\alpha\bigr)}{\alpha^{2}} = -\frac{2x}{\alpha}, \] positive for $x\lt 0$ and negative for $x\gt 0$: a local maximum.

Step 5: Verification

\[\alpha = 1:\ \text{the cubic gives } x = 2 \ (x = -1 \text{ is the pole}),\ y = \frac{8+1}{4-1} = 3 = \tfrac32\cdot2 \;\;\checkmark\]

\[\alpha = 4:\ x = -3.064178,\ -0.694593,\ 3.758771 \ \Rightarrow\ y = -4.596267,\ -1.041889,\ 5.638156 = \tfrac32x \;\;\checkmark\]

\[\alpha = 0.3:\ x = 1.185752 \Rightarrow y = 1.778628 = \tfrac32x; \qquad\text{and } y(0) = -1 \text{ for every } \alpha \;\;\checkmark\]