Помогите, Найти точку минимума функции.

$y= \frac{1}{3} x^{3} -2.5 x^{2} +4x\\\\f'(x)= x^{2} -5x+4\\f'(x)=0\\ x^{2} -5x+4=0\\D=25-16=9\\ \sqrt{D} =3\\ x_{1} = \frac{5+3}{2} =4\\ x_{2} = \frac{5-3}{2} =1 \\ x^{2} -5x+4=(x-1)(x-4)$
f'(x) + max - min +
------------------|---------------------------|----------------------------> x
f(x) 1 4

$f _{min} =f(4)= ( \frac{1}{3})*4 ^{3}-2.5*4 ^{2} +4*4= \\ = \frac{1}{3} *64-2.5*16+16= \frac{64}{3} -40+16= \\ = \frac{64}{3} -24= \frac{64-72}{3} = \frac{8}{3} =2 \frac{2}{3}$