Question

[Chú Cứ Để Anh]

Tìm x để y' > 0 

y = x + 1 + 1/x+1 

Answer 1

[nguyen quynh nhu]

ko nho

Answer 2

[Chú Cứ Để Anh]

\(y=x+1+\frac{1}{x+1}\left(Đk:x\ne-1\right)\)

\(\rightarrow y'=1+0+\frac{1'.\left(x+1\right)-1.\left(x+1\right)'}{\left(x+1\right)^2}\)

     \(y'=1+\frac{-1}{\left(x+1\right)^2}\)

    \(y'=1-\frac{1}{\left(x+1\right)^2}\)

   \(y'=\frac{x^2+2x+1-1}{\left(x+1\right)^2}\)

   \(y'=\frac{x^2+2x}{\left(x+1\right)^2}\)

Để y' > 0 \(\Leftrightarrow\frac{x^2+2x}{\left(x+1\right)^2}>0\)

                Mà \(\left(x+1\right)^2>0\)

       \(\rightarrow x^2+2x>0\)

       \(\Leftrightarrow\orbr{\begin{cases}x< -2\\x>0\end{cases}}\)