$\displaystyle y = {x^{\frac{2}
{3}}}(x - 4)$
What is the maximum value that this function takes on the interval [0,5]. (not plug and chug)
$\displaystyle y = x^{\frac{5}{3}} - 4x^{\frac{2}{3}}$
$\displaystyle y' = \frac{5}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}}
$
note that x = 0 is a critical value
set y' = 0
$\displaystyle \frac{5}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}} = 0$
multiply every term by $\displaystyle 3x^{\frac{1}{3}}$ ...
$\displaystyle 5x - 8 = 0$
$\displaystyle x = \frac{8}{5}$
for $\displaystyle 0 < x < \frac{8}{5}$ , y'(x) < 0 ... y is decreasing
for $\displaystyle \frac{8}{5} < x < 5$ , y'(x) > 0 ... y is increasing
since the function decreases, then increases, that says the function has maxima at the endpoints. comparing the function values at the endpoints ...
$\displaystyle y(0) = 0$
$\displaystyle y(5) = 5^{\frac{2}{3}} > 0$
$\displaystyle y(5) = 5^{\frac{2}{3}}$ is the maximum over the interval.