Объяснение:

[tex]f(x)=\frac{2x^3}{3} -8x \ \ \ \ \ x\in[0;3].\\f'(x)=(\frac{2x^3}{3}-8x)'=2x^2-8=0\\ 2x^2-8=0\ |:2\\x^2-4=0\\x^2-2^2=0\\(x+2)*(x-2)=0\\x_1=-2\notin\\x_2=2\in.\\f(0)=\frac{2*0^3}{3}-8*0=0-0=0.\\ f(2)=\frac{2*2^3}{3} -8*2=\frac{2*8}{3} -16=\frac{16}{3} -16=5\frac{1}{3}-16=-10\frac{2}{3} .\\ f(3)=\frac{2*3^3}{3} -8*3=2*9-24=18-24=-6.[/tex]

Ответ: f(0)=0=yнаиб.