Find the inflection points of the curve y = 3x^5-2x^6
Find the inflection points of the curve $y = 3x^{5}-2x^{6}$.
Explanation of Method
An odd power at the origin, so the concavity does change.
Step 1: The derivatives
\[ y' = 15x^{4}-12x^{5}, \qquad y'' = 60x^{3}-60x^{4} = 60x^{3}\bigl(1-x\bigr). \]
Step 2: Candidates
\[ y'' = 0 \;\Longleftrightarrow\; x = 0 \ \text{ or } \ x = 1. \]
Step 3: Sign of $y''$
\[ x\lt 0:\ \frac{(-)(+)}{} = -; \qquad 0\lt x\lt 1:\ (+)(+) = +; \qquad x\gt 1:\ (+)(-) = -. \]
The cube $x^{3}$ changes sign at the origin — an odd power — so $y''$ really does change sign there, and again at $x = 1$ through the factor $1-x$.
Step 4: The two points
\[ y(0) = 0, \qquad y(1) = 3-2 = 1. \]
Step 5: Verification
\[y''(-1) = -120\lt 0; \qquad y''(0.5) = 3.75\gt 0; \qquad y''(2) = -480\lt 0 \;\;\checkmark\]
\[y'(0) = 0 \text{ as well: the origin is a stationary inflection, with a horizontal tangent} \;\;\checkmark\]
\[y'(1) = 15-12 = 3\neq0:\ \text{at } \bigl(1,1\bigr) \text{ the tangent is oblique} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\bigl(0,\,0\bigr) \qquad\text{and}\qquad \bigl(1,\,1\bigr)\,}\]
Remark
Remark. Had the exponents been even — $y = 3x^{4}-2x^{6}$, say — the origin would have been an extremum instead, since $y''$ would keep its sign. It is the odd multiplicity of the root $x = 0$ in $y''$ that makes it an inflection, exactly as in Problem 431.