Question

Найти наибольшее и наименьшее значение функции $F(x) = x-frac{4}{5} x^{3}$ на промежутке [0,1]

Answer 1

$f(x)=x-frac{4}{5}, x^3; ; ,; ; xin [, 0,1, ]\\f'(x)=1-frac{4}{5}cdot 3x^2=1-frac{12}{5}, x^2=0; ; Rightarrow ; ; ; frac{5-12x^2}{5}=0; ,\\x^2=frac{5}{12}; ,; ; x_{1.2}=pm sqrt{frac{5}{12}}\\x_1=-sqrt{frac{5}{12}}notin [, 0,1, ]\\fleft (sqrt{frac{5}{12}}right )=sqrt{frac{5}{12}}-frac{4}{5}cdot sqrt{frac{5^3}{12^3}}=sqrt{frac{5}{12}}-frac{4cdot 5sqrt5}{5cdot 12sqrt{12}}=sqrt{frac{5}{12}}-frac{1}{3}cdot sqrt{frac{5}{12}}=frac{2}{3}cdot sqrt{frac{5}{12}}$

$f(0)=0\\f(1)=1-frac{4}{5}=frac{1}{5}=0,2\\f(naibol)=fleft (sqrt{frac{5}{12}}right )=frac{2}{3}cdot sqrt{frac{5}{12}} ; ; ; ,; ; f(naimen)=f(0)=0$