tex] на промежутке [0,1]...

July 2022 1 5 Report

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

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

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

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