a) Với bất kì $x_{0}: f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}$

Q: a) Với bất kì $x_{0}: f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}$

A. True B. False Với bất kì $x_{0}$, ta có:$\begin{aligned} f^{\prime}\left(x_{0}\right) & =\lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}=\lim _{x \rightarrow x_{0}} \frac{2 x^{3}-2 x_{0}^{3}}{x-x_{0}} \\ & =\lim _{x \rightarrow x_{0}} \frac{2\left(x-x_{0}\right)\left(x^{2}+x \cdot x_{0}+x_{0}^{2}\right)}{x-x_{0}} \\ & =\lim _{x \rightarrow x_{0}} \frac{2\left(x^{2}+x \cdot x_{0}+x_{0}^{2}\right)}{1}=6 x_{0}^{2} .\end{aligned}$Vậy $f^{\prime}(x)=\left(2 x^{3}\right)^{\prime}=6 x^{2}$ trên $\mathbb{R}$.

Question

Accepted Answer

A. True
B. False

Với bất kì $x_{0}$, ta có:

$\begin{aligned} f^{\prime}\left(x_{0}\right) & =\lim _{x \rightarrow x_{0}} \frac{f(x)-f\left(x_{0}\right)}{x-x_{0}}=\lim _{x \rightarrow x_{0}} \frac{2 x^{3}-2 x_{0}^{3}}{x-x_{0}} \ & =\lim _{x \rightarrow x_{0}} \frac{2\left(x-x_{0}\right)\left(x^{2}+x \cdot x_{0}+x_{0}^{2}\right)}{x-x_{0}} \ & =\lim _{x \rightarrow x_{0}} \frac{2\left(x^{2}+x \cdot x_{0}+x_{0}^{2}\right)}{1}=6 x_{0}^{2} .\end{aligned}$

Vậy $f^{\prime}(x)=\left(2 x^{3}\right)^{\prime}=6 x^{2}$ trên $\mathbb{R}$.

Đề thi giữa kì 2 (Cấu trúc mới) - Đề số 27 - MĐ 10958