Question

Cho hàm số \(f\left(x\right)\left\{{}\begin{matrix}\dfrac{3x^2+ax+3}{x-3}\forall x>3\\x^2+bx+c\forall x\le3\end{matrix}\right.\) có đạo hàm tại 3

Tìm a,b,c 

Answer 1

\(3.3^2+3a+3=0\Rightarrow a=-10\)

\(\lim\limits_{x\rightarrow3}\dfrac{3x^2-10x+3}{x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(3x-1\right)}{x-3}=\lim\limits_{x\rightarrow3}\left(3x-1\right)=8\)

\(\lim\limits_{x\rightarrow3^-}\left(x^2+bx+c\right)=3b+c+9\)

\(\Rightarrow3b+c+9=8\)

\(f'\left(3^+\right)=\left(3x-1\right)'=3\)

\(f'\left(3^-\right)=6+b\Rightarrow6+b=3\Rightarrow b=-3\)

\(\Rightarrow c=8\)