Question

tìm a,b

f(x)= \(\left\{{}\begin{matrix}2x^2+ax+b\text{ khi x< -1}\\4x+2a\text{ khi -1\le x< 1}\\ax^2+bx\text{ khi x}\ge1\end{matrix}\right.\) có giới hạn khi x=-1 và x=1

Answer 1

\(\lim\limits_{x\rightarrow-1^-}f\left(x\right)=2-a+b\)

\(\lim\limits_{x\rightarrow-1^+}f\left(x\right)=-4+2a\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=4+2a\)

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=a+b\)

Để hs có giới hạn tại \(x=1;-1\) thì:

\(\left\{{}\begin{matrix}2-a+b=-4+2a\\4+2a=a+b\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3a-b=2\\a-b=-4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3\\b=7\end{matrix}\right.\)

Answer 2

4x+2a khi -1 <=x<1 nha mn