Show that all the inflection points of the curve y = x
Show that all the inflection points of the curve $y = x\sin x$ lie on the curve $x^{2}y^{2} = 4\bigl(x^{2}-y^{2}\bigr)$.
Explanation of Method
The Pythagorean identity does the elimination.
Step 1: The second derivative
\[ y' = \sin x+x\cos x, \qquad y'' = 2\cos x-x\sin x. \tag{1} \]
Step 2: The inflection condition
At an inflection $y'' = 0$, that is
\[ x\sin x = 2\cos x. \tag{2} \]
Note first that $\sin x\neq0$ at such a point: $\sin x = 0$ would force $\cos x = 0$ by (2), which is impossible.
Step 3: Read off the sine and the cosine
If $\bigl(x,y\bigr)$ is an inflection point then $y = x\sin x$, and (2) says this same quantity equals $2\cos x$. Hence
\[ \sin x = \frac{y}{x}, \qquad \cos x = \frac{y}{2}. \tag{3} \]
Step 4: Eliminate the angle
Substituting (3) into $\sin^{2}x+\cos^{2}x = 1$: $\frac{y^{2}}{x^{2}}+\frac{y^{2}}{4} = 1$.
Multiplying by $4x^{2}$ and rearranging, $4y^{2}+x^{2}y^{2} = 4x^{2} \;\Longleftrightarrow\; x^{2}y^{2} = 4\bigl(x^{2}-y^{2}\bigr)$.
Step 5: Verification
\[\text{First inflection: } x = 1.0768740,\ y = 0.9481661:\quad x^{2}y^{2} = 1.0425542 = 4\bigl(x^{2}-y^{2}\bigr) \;\;\checkmark\]
\[x = 3.6435972,\ y = -1.7532396:\quad \text{both sides } = 40.807805 \;\;\checkmark\]
\[x = 6.5783337,\ y = 1.9135180:\quad \text{both sides } = 158.45169 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\text{every inflection point of } y = x\sin x \text{ satisfies } x^{2}y^{2} = 4\bigl(x^{2}-y^{2}\bigr)\,}\]
Remark
Remark. The inflection abscissae themselves satisfy the transcendental equation $\cot x = \dfrac{x}{2}$ and cannot be written in closed form — yet the locus they trace is an ordinary algebraic curve. That is the usual pattern: eliminating a transcendental unknown through an identity it must satisfy leaves an algebraic relation between the coordinates. Solving for $y$ gives $y^{2} = \dfrac{4x^{2}}{x^{2}+4}$, so $|y|\lt 2$: all inflections lie in the horizontal strip $|y|\lt 2$.