Решите( с подробным решением),заранее большое спасибо
спасибо большое
А как понять [tex]?
спасибо огромнейшее
![f(x)= \frac{4}{x+1} +x \\ f'(x)=( \frac{4}{x+1})'+(x)'= \frac{(4)'(x+1)-4(x+1)'}{(x+1)^2} +1=[tex] \frac{-4}{(x+1)^2}=-1 \\ \frac{4}{(x+1)^2}=1 \\ \\ (x+1)^2=4 \\ x+1=2 \\ x=1 \\ \\ x+1=-2 \\ x=-3](https://tex.z-dn.net/?f=f%28x%29%3D+%5Cfrac%7B4%7D%7Bx%2B1%7D+%2Bx+%5C%5C+f%27%28x%29%3D%28+%5Cfrac%7B4%7D%7Bx%2B1%7D%29%27%2B%28x%29%27%3D++%5Cfrac%7B%284%29%27%28x%2B1%29-4%28x%2B1%29%27%7D%7B%28x%2B1%29%5E2%7D+%2B1%3D%5Btex%5D+%5Cfrac%7B-4%7D%7B%28x%2B1%29%5E2%7D%3D-1+%5C%5C++%5Cfrac%7B4%7D%7B%28x%2B1%29%5E2%7D%3D1+%5C%5C+%5C%5C++%28x%2B1%29%5E2%3D4+%5C%5C+x%2B1%3D2+%5C%5C+x%3D1+%5C%5C++%5C%5C+x%2B1%3D-2+%5C%5C+x%3D-3 "f(x)= \frac{4}{x+1} +x \\ f'(x)=( \frac{4}{x+1})'+(x)'= \frac{(4)'(x+1)-4(x+1)'}{(x+1)^2} +1=[tex] \frac{-4}{(x+1)^2}=-1 \\ \frac{4}{(x+1)^2}=1 \\ \\ (x+1)^2=4 \\ x+1=2 \\ x=1 \\ \\ x+1=-2 \\ x=-3")
![[0;3]](https://tex.z-dn.net/?f=%5B0%3B3%5D "[0;3]")
Ответ: 