Finding critical numbers with fractional exponents

Differentiate: $f'(x) = \frac{4}{3}x^{1/3} - \frac{4}{3}x^{-2/3}$. Factor out $\frac{4}{3}x^{-2/3}$ to get $f'(x) = \frac{4}{3}x^{-2/3}(x - 1) = \frac{4(x-1)}{3x^{2/3}}$. This form makes it easy to see where the derivative is zero or undefined.

The derivative equals zero when the numerator $4(x-1) = 0$, giving $x = 1$. The derivative is undefined when the denominator $3x^{2/3} = 0$, giving $x = 0$; note $f(0) = 0$ is defined, so $x=0$ qualifies as a critical number. Both $x = 0$ and $x = 1$ lie within $[-1,5]$, so these are the critical numbers.