Find the curve on which all the inflection points of y =

Find the curve on which all the inflection points of $y = \dfrac{\sin x}{x}$, $x\neq0$, lie.

Step 1: The derivatives

\[ y' = \frac{x\cos x-\sin x}{x^{2}}, \] and differentiating again, with $\bigl(x\cos x-\sin x\bigr)' = -x\sin x$: \[y'' = \frac{-x^{3}\sin x-2x\bigl(x\cos x-\sin x\bigr)}{x^{4}} = \frac{\bigl(2-x^{2}\bigr)\sin x-2x\cos x}{x^{3}}. \tag{1}\]

Step 2: The inflection condition

\[ \bigl(2-x^{2}\bigr)\sin x = 2x\cos x. \tag{2} \]

Here too $\sin x\neq0$ at an inflection: $\sin x = 0$ would give $x = k\pi$ and then (2) reads $-2k\pi\bigl(-1\bigr)^{k} = 0$, which is false.

Step 3: Read off the sine and the cosine

From $y = \dfrac{\sin x}{x}$ we get $\sin x = xy$, and substituting into (2), \[2x\cos x = \bigl(2-x^{2}\bigr)xy \;\Longrightarrow\; \cos x = \frac{\bigl(2-x^{2}\bigr)y}{2}. \tag{3}\]

Step 4: Eliminate the angle

\[ \sin^{2}x+\cos^{2}x = x^{2}y^{2}+\frac{\bigl(2-x^{2}\bigr)^{2}y^{2}}{4} = 1, \] and since $4x^{2}+\bigl(2-x^{2}\bigr)^{2} = 4x^{2}+4-4x^{2}+x^{4} = x^{4}+4$, $\frac{y^{2}\bigl(x^{4}+4\bigr)}{4} = 1 \;\Longleftrightarrow\; y^{2}\bigl(x^{4}+4\bigr) = 4$.

Step 5: Verification

\[\text{First inflection: } x = 2.0815760,\ y = 0.4190881:\quad y^{2}\bigl(x^{4}+4\bigr) = 4.0000000 \;\;\checkmark\]

\[x = 5.9403700,\ y = -0.0565857:\quad y^{2}\bigl(x^{4}+4\bigr) = 4.0000000 \;\;\checkmark\]

\[x = 9.2058401,\ y = 0.0235929:\quad y^{2}\bigl(x^{4}+4\bigr) = 4.0000000 \;\;\checkmark\]