Cho hàm số \(f\left(x\right)=\left\{{}\begin{matrix}x^2+2x+1;\left(x\ge0\right)\\2x+1;\left(x< 0\right)\end{matrix}\right.\). Tính \(f'\left(0\right)\) ?
\(\dfrac{1}{4}\) \(2\) \(-2\) Không tồn tại Hướng dẫn giải:\(f\left(x\right)=\left\{{}\begin{matrix}x^2+2x+1;\left(x\ge0\right)\\2x+1;\left(x< 0\right)\end{matrix}\right.\)\(\Rightarrow\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\left(x^2+2x+1\right)=1;\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\left(2x+1\right)=1\)\(\Rightarrow\lim\limits_{x\rightarrow0}f\left(x\right)=f\left(0\right)\) nên hàm số liên tục tại \(x=0\).
Hơn nữa \(f'\left(0^+\right)=\lim\limits_{t\left(>0\right)\rightarrow0}\dfrac{f\left(0+t\right)-f\left(0\right)}{t}=\lim\limits_{t\rightarrow0}\dfrac{t^2+2t}{t}=\lim\limits_{t\rightarrow0}\left(t+2\right)=2\)
và \(f'\left(0^-\right)=\lim\limits_{t\left(< 0\right)\rightarrow0}\dfrac{f\left(0+t\right)-f\left(0\right)}{t}=\lim\limits_{t\rightarrow0}\dfrac{2t}{t}=\lim\limits_{t\rightarrow0}\left(2\right)=2\), suy ra \(f'\left(0\right)=2\)