Examine the sign of the second derivative of y = fourth

Examine the sign of the second derivative of $y = \sqrt[4]{2x^{2}-x^{3}}$.

Step 1: Domain

The radicand $u = 2x^{2}-x^{3} = x^{2}\bigl(2-x\bigr)$ must be non-negative, which for $x^{2}\ge0$ means $2-x\ge0$: \[\text{domain } = \bigl(-\infty,\,2\bigr], \qquad y = 0 \text{ exactly at } x = 0 \text{ and } x = 2.\]

Step 2: A convenient formula for $y''$

For $y = u^{1/4}$, \[y' = \frac14u^{-3/4}u', \qquad y'' = \frac{1}{16}\,u^{-7/4}\left[4uu''-3\bigl(u'\bigr)^{2}\right]. \tag{1}\]

Here $u' = 4x-3x^{2}$ and $u'' = 4-6x$.

Step 3: The bracket

\[ 4uu'' = 4\bigl(2x^{2}-x^{3}\bigr)\bigl(4-6x\bigr) = 32x^{2}-64x^{3}+24x^{4}, \]

\[ 3\bigl(u'\bigr)^{2} = 3x^{2}\bigl(4-3x\bigr)^{2} = 48x^{2}-72x^{3}+27x^{4}, \]

\[ 4uu''-3\bigl(u'\bigr)^{2} = -16x^{2}+8x^{3}-3x^{4} = -x^{2}\bigl(3x^{2}-8x+16\bigr). \]

Hence \[y'' = \frac{-x^{2}\bigl(3x^{2}-8x+16\bigr)}{16\bigl(2x^{2}-x^{3}\bigr)^{7/4}}. \tag{2}\]

Step 4: Every factor has a fixed sign

The quadratic $3x^{2}-8x+16$ has discriminant $64-192 = -128\lt 0$ and positive leading coefficient, so it is positive for every real $x$. Also $-x^{2}\lt 0$ for $x\neq0$, and $\bigl(2x^{2}-x^{3}\bigr)^{7/4}\gt 0$ on the interior of the domain. Therefore

\[ y''\lt 0 \qquad\text{for every } x\in\bigl(-\infty,0\bigr)\cup\bigl(0,2\bigr). \]

Step 5: Consequence

The second derivative never changes sign, so the curve is concave downward on the whole of its domain and has no inflection points. At $x = 0$ and $x = 2$ the derivative is infinite and $y''$ undefined; these are cusp-like ends, not inflections.

Step 6: Verification

\[3x^{2}-8x+16 = 3\left(x-\tfrac43\right)^{2}+\tfrac{32}{3} \ge \tfrac{32}{3} \gt 0 \;\;\checkmark\]

\[y(1) = 1,\ y(1.5) = \sqrt[4]{1.125} = 1.0299,\ y(0.5) = \sqrt[4]{0.375} = 0.7825:\ \text{a single hump} \;\;\checkmark\]

\[\text{Maximum where } u' = 0,\ \text{i.e. } x = \tfrac43:\ y = \sqrt[4]{\tfrac{32}{27}} = 1.0434 \;\;\checkmark\]