Применено определение процента
Найдем стационарные точки:
![f(x)=\frac{x^{2}+7x}{x-9}\\\\ f'(x)=[\frac{x^{2}+7x}{x-9} ]'=\frac{[x^2+7x]'*[x-9]-[x^2+7x]*[x-9]'}{(x-9)^2}=\\\\ =\frac{[2x+7]*[x-9]-[x^2+7x]*[1]}{(x-9)^2}=\frac{2x^2-18x+7x-63-x^2-7x}{(x-9)^2}=\\\\ =\frac{x^2-18x-63}{(x-9)^2}.\\\\ f'(x)=0\\\\ \frac{x^2-18x-63}{(x-9)^2}=0\\\\ \frac{x^2-21x+3x-63}{(x-9)^2}=0\\\\ \frac{x(x-21)+3(x-21)}{(x-9)^2}=0\\\\ \frac{(x+3)(x-21)}{(x-9)^2}=0\\\\ x_1=-3\ \ x_2=21](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx%5E%7B2%7D%2B7x%7D%7Bx-9%7D%5C%5C%5C%5C%0Af%27%28x%29%3D%5B%5Cfrac%7Bx%5E%7B2%7D%2B7x%7D%7Bx-9%7D+%5D%27%3D%5Cfrac%7B%5Bx%5E2%2B7x%5D%27%2A%5Bx-9%5D-%5Bx%5E2%2B7x%5D%2A%5Bx-9%5D%27%7D%7B%28x-9%29%5E2%7D%3D%5C%5C%5C%5C%0A%3D%5Cfrac%7B%5B2x%2B7%5D%2A%5Bx-9%5D-%5Bx%5E2%2B7x%5D%2A%5B1%5D%7D%7B%28x-9%29%5E2%7D%3D%5Cfrac%7B2x%5E2-18x%2B7x-63-x%5E2-7x%7D%7B%28x-9%29%5E2%7D%3D%5C%5C%5C%5C%0A%3D%5Cfrac%7Bx%5E2-18x-63%7D%7B%28x-9%29%5E2%7D.%5C%5C%5C%5C%0Af%27%28x%29%3D0%5C%5C%5C%5C%0A%5Cfrac%7Bx%5E2-18x-63%7D%7B%28x-9%29%5E2%7D%3D0%5C%5C%5C%5C%0A%5Cfrac%7Bx%5E2-21x%2B3x-63%7D%7B%28x-9%29%5E2%7D%3D0%5C%5C%5C%5C%0A%5Cfrac%7Bx%28x-21%29%2B3%28x-21%29%7D%7B%28x-9%29%5E2%7D%3D0%5C%5C%5C%5C%0A%5Cfrac%7B%28x%2B3%29%28x-21%29%7D%7B%28x-9%29%5E2%7D%3D0%5C%5C%5C%5C%0Ax_1%3D-3%5C+%5C+x_2%3D21)
![f'(x)=\frac{(x+3)(x-21)}{(x-9)^2}\\\\ +++++[-3]------(9)-----[21]++++\ \textgreater \ x](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B%28x%2B3%29%28x-21%29%7D%7B%28x-9%29%5E2%7D%5C%5C%5C%5C%0A%2B%2B%2B%2B%2B%5B-3%5D------%289%29-----%5B21%5D%2B%2B%2B%2B%5C+%5Ctextgreater+%5C+x)
Получили, что при значении
![x=-3](https://tex.z-dn.net/?f=x%3D-3)
функция
![f(x)](https://tex.z-dn.net/?f=f%28x%29)
достигает своего максимума:
![f(-3)=\frac{(-3)^{2}+7*(-3)}{-3-9} =\frac{9-21}{-12}=1](https://tex.z-dn.net/?f=f%28-3%29%3D%5Cfrac%7B%28-3%29%5E%7B2%7D%2B7%2A%28-3%29%7D%7B-3-9%7D+%3D%5Cfrac%7B9-21%7D%7B-12%7D%3D1)
также, при значении
![x=21](https://tex.z-dn.net/?f=x%3D21)
функция
![f(x)](https://tex.z-dn.net/?f=f%28x%29)
достигает своего максимума:
![f(21)=\frac{21^{2}+7*21}{21-9}=\frac{588}{12}=49\\\\ ](https://tex.z-dn.net/?f=f%2821%29%3D%5Cfrac%7B21%5E%7B2%7D%2B7%2A21%7D%7B21-9%7D%3D%5Cfrac%7B588%7D%7B12%7D%3D49%5C%5C%5C%5C%0A)
на концах интервала значения функции:
![f(-4)=\frac{(-4)^{2}+7*(-4)}{-4-9} =\frac{16-28}{-13}=\frac{12}{13}\\\\ f(1)=\frac{1^{2}+7*1}{1-9} =\frac{8}{-8}=-1](https://tex.z-dn.net/?f=f%28-4%29%3D%5Cfrac%7B%28-4%29%5E%7B2%7D%2B7%2A%28-4%29%7D%7B-4-9%7D+%3D%5Cfrac%7B16-28%7D%7B-13%7D%3D%5Cfrac%7B12%7D%7B13%7D%5C%5C%5C%5C%0Af%281%29%3D%5Cfrac%7B1%5E%7B2%7D%2B7%2A1%7D%7B1-9%7D+%3D%5Cfrac%7B8%7D%7B-8%7D%3D-1)
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В итоге, наибольшее значение функции на промежутке
![[-4;\ 1]](https://tex.z-dn.net/?f=%5B-4%3B%5C+1%5D)
равно
![f(-3)=1](https://tex.z-dn.net/?f=f%28-3%29%3D1)
, и наименьшее:
![f(1)=-1](https://tex.z-dn.net/?f=f%281%29%3D-1)
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Ответ: на промежутке
![-1 \leq f(x) \leq 1](https://tex.z-dn.net/?f=-1+%5Cleq+f%28x%29+%5Cleq+1)
5 1/3-2=3 1/3
3 1/3:5/21=10/3*21/5=14
если я правильно понимаю что надо сделать
Yo=cos(Pi/3-4*pi/8)=cos(-pi/6)=root(3)/2
Yo=tg(3Pi/12-Pi/4)=tg(0)=0