Locating and classifying a cusp using the derivative

Rewrite $v(x) = x^{5/3} - 5x^{2/3}$, so $v'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{-1/3}(x-2)$. This is undefined at $x=0$ because of the $x^{-1/3}$ factor, while $v(0)=0$ is defined, so $x=0$ is a critical point where the tangent line is vertical — a candidate for a cusp.

For $x$ slightly less than 0 (e.g., $x = -0.1$), $x^{-1/3} < 0$ (the real cube root of a negative number is negative) and $(x-2) < 0$, so $v'(x) = \frac{5}{3}x^{-1/3}(x-2) > 0$. For $x$ slightly greater than 0 (e.g., $x = 0.1$), $x^{-1/3} > 0$ and $(x-2) < 0$, so $v'(x) < 0$. Since $v'$ changes from positive to negative as $x$ passes through 0, and $v'$ approaches $\pm\infty$ on each side (creating a vertical tangent from both directions), $x=0$ is a cusp. The sign change from $v' > 0$ to $v' < 0$ means $v$ increases before 0 and decreases after 0, making $x=0$ a local maximum.

The tangent line at the cusp is vertical. This is because the derivative involves a factor with a negative one-third power of x, which blows up to positive or negative infinity as x approaches zero from either side, rather than approaching a finite slope.