Ответы
Ответ дал:
2
D(y)=(0;+∞)
![y'=6- \frac{6}{6x} =6- \frac{1}{x} = \frac{6x-1}{x} \\ \\ y'=0\ npu\ x= \frac{1}{6} \in [ \frac{1}{12}; \frac{5}{12} ] \\ \\ y(\frac{1}{6} )=6*\frac{1}{6} -ln(6*\frac{1}{6} )+17=1-0+17=18 \\ \\
y(\frac{1}{12} )=6*\frac{1}{12} -ln(6*\frac{1}{12} )+17=0,5+ln2+17=17,5+ln2\ \textgreater \ 18,\\ m.k.\ ln2\ \textgreater \ 0,5 \\ \\ y'=6- \frac{6}{6x} =6- \frac{1}{x} = \frac{6x-1}{x} \\ \\ y'=0\ npu\ x= \frac{1}{6} \in [ \frac{1}{12}; \frac{5}{12} ] \\ \\ y(\frac{1}{6} )=6*\frac{1}{6} -ln(6*\frac{1}{6} )+17=1-0+17=18 \\ \\
y(\frac{1}{12} )=6*\frac{1}{12} -ln(6*\frac{1}{12} )+17=0,5+ln2+17=17,5+ln2\ \textgreater \ 18,\\ m.k.\ ln2\ \textgreater \ 0,5 \\ \\](https://tex.z-dn.net/?f=y%27%3D6-+%5Cfrac%7B6%7D%7B6x%7D+%3D6-+%5Cfrac%7B1%7D%7Bx%7D+%3D+%5Cfrac%7B6x-1%7D%7Bx%7D++%5C%5C+%5C%5C+y%27%3D0%5C+npu%5C+x%3D+%5Cfrac%7B1%7D%7B6%7D+%5Cin+%5B+%5Cfrac%7B1%7D%7B12%7D%3B+%5Cfrac%7B5%7D%7B12%7D++%5D+%5C%5C+%5C%5C+y%28%5Cfrac%7B1%7D%7B6%7D+%29%3D6%2A%5Cfrac%7B1%7D%7B6%7D+-ln%286%2A%5Cfrac%7B1%7D%7B6%7D+%29%2B17%3D1-0%2B17%3D18+%5C%5C++%5C%5C+%0Ay%28%5Cfrac%7B1%7D%7B12%7D+%29%3D6%2A%5Cfrac%7B1%7D%7B12%7D+-ln%286%2A%5Cfrac%7B1%7D%7B12%7D+%29%2B17%3D0%2C5%2Bln2%2B17%3D17%2C5%2Bln2%5C+%5Ctextgreater+%5C+18%2C%5C%5C+m.k.%5C+ln2%5C+%5Ctextgreater+%5C+0%2C5+%5C%5C+%5C%5C)
![y(\frac{5}{12} )=6*\frac{5}{12} -ln(6*\frac{5}{12} )+17=2,5-ln2,5+17= \\ =19,5-ln2,5\ \textgreater \ 18, \ m.k.\ ln2,5\ \textless \ 1. y(\frac{5}{12} )=6*\frac{5}{12} -ln(6*\frac{5}{12} )+17=2,5-ln2,5+17= \\ =19,5-ln2,5\ \textgreater \ 18, \ m.k.\ ln2,5\ \textless \ 1.](https://tex.z-dn.net/?f=y%28%5Cfrac%7B5%7D%7B12%7D+%29%3D6%2A%5Cfrac%7B5%7D%7B12%7D+-ln%286%2A%5Cfrac%7B5%7D%7B12%7D+%29%2B17%3D2%2C5-ln2%2C5%2B17%3D+%5C%5C+%3D19%2C5-ln2%2C5%5C+%5Ctextgreater+%5C+18%2C+%5C+m.k.%5C+ln2%2C5%5C+%5Ctextless+%5C+1.)
Итак, наим.значение функции на заданном отрезке - число 18
Итак, наим.значение функции на заданном отрезке - число 18
Вас заинтересует
10 месяцев назад
10 месяцев назад
1 год назад
5 лет назад