Question

f(x)=x$\sqrt{1-x^2}$ Найти наибольшее значение функции на интервале (0;1)

Answer 1

$\displaystyle\\f(x)=x\sqrt{1-x^2}\\\\f'(x)=x'\cdot \sqrt{1-x^2}+x\cdot (\sqrt{1-x^2})'=\sqrt{1-x^2}+x\cdot(-2x)\cdot\frac{1}{2\sqrt{1-x^2}}\\\\f'(x)=0:\\\\\\\sqrt{1-x^2}-\frac{2x^2}{2\sqrt{1-x^2}}=0\\\\\\ \sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=0\\\\\\\frac{1-2x^2}{\sqrt{1-x^2}}=0\\\\\\ODZ:1-x^2>0\\\\-x^2>-1\\\\x\in(-1;1)\\\\1-2x^2=0\\\\-2x^2=-1\\\\x^2=\frac{1}{2}\\\\x=\pm\sqrt{\frac{1}{2} }=\pm\frac{1}{\sqrt{2}}$

$\displaystyle\\-\frac{1}{\sqrt{2}}\notin (0;1)\\\\\\ f\bigg(\frac{1}{\sqrt{2}}\bigg)=\frac{1}{\sqrt{2}}\sqrt{1-\bigg(\frac{1}{\sqrt{2}}\bigg)^2 } =\frac{1}{\sqrt{2}}\sqrt{1-\frac{1}{2 } } =\frac{1}{\sqrt{2}}\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\cdot \frac{1}{\sqrt{2}}=\\\\\\ =\frac{\sqrt{2}}{2\sqrt{2}}=\frac{1}{2} \\\\\\f_{max\ (0;1)}=\frac{1}{2}$