A student plots the graph of f(x) = x^(2/3)(x - 5) by computer and wants to find its local extreme…

Step 1. The derivative. $f(x) = x^{5/3} - 5x^{2/3}$, so $f'(x) = \dfrac{5}{3}x^{2/3} - \dfrac{10}{3}x^{-1/3} = \dfrac{5(x - 2)}{3x^{1/3}}$.

Step 2. The critical points. $f'(x) = 0$ at $x = 2$. The derivative does not exist at $x = 0$, where the denominator is $0$, but $f(0) = 0$ is defined. So $x = 0$ is a critical point too.

Step 3. The signs. At $x = -1$: $f' = \dfrac{5(-3)}{3(-1)} = 5 > 0$. At $x = 1$: $f' = \dfrac{5(-1)}{3} < 0$. At $x = 3$: $f' > 0$.

Step 4. The kinds. At $x = 0$, $f'$ goes from $+$ to $-$, a local maximum with value $f(0) = 0$. At $x = 2$, $f'$ goes from $-$ to $+$, a local minimum with value $f(2) = 2^{2/3} \times (-3) = -4.762$.

Step 5. The second derivative test. It needs $f'(c) = 0$ and $f''(c)$ to exist. At $x = 0$, $f'$ does not exist, so neither does $f''$.

The shape. Near $x = 0$ the slope is large and positive on the left and large and negative on the right, so the graph comes to a sharp tip that points upwards. This is called a cusp.