Cho hàm số $f\left(x\right)=\left\{\begin{array}{l}{\left(x-1\right)}^{2}khix\ge 0\\ 1-2xkhix<0\end{array}\right.$ . Tính f'(0).

Ta có f(0) = (0 – 1)² = 1.

Ta có $\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}$$=\underset{x\to 0^{+}}{\lim }\frac{{\left(x-1\right)}^2-1}{x}$$=\underset{x\to 0^{+}}{\lim }\frac{x^2-2x}{x}$$=\underset{x\to 0^{+}}{\lim }\left(x-2\right)=-2$

$\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}$$=\underset{x\to 0^{-}}{\lim }\frac{1-2x-1}{x}$$=\underset{x\to 0^{-}}{\lim }\frac{-2x}{x}=-2$

Suy ra $\underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=-2$

Vậy f'(0) = −2.