Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)
a=?
b=?
Can someone show me the steps on how to do it? I think i messed up in substituting (-2) in both equations.
Thank you in advance
Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)
a=?
b=?
Can someone show me the steps on how to do it? I think i messed up in substituting (-2) in both equations.
Thank you in advance
The answer to the question in red is no.
That is not the object of this site. We want you to learn something.
So here is some help.
If $\displaystyle \lim _{x \to 1^ - } f(x) = 2$ then what are $\displaystyle a~\&~b$ so that $\displaystyle \lim _{x \to 1^ + } f(x) = 2?$
$\displaystyle \lim_{x\to1^-}\frac{x^2-1}{x-1}=\lim_{x\to1^+}(ax^2+bx+2)$
Which implies
1. $\displaystyle 0=a+b$ or $\displaystyle a=-b$
$\displaystyle \lim_{x\to3^+}(3x-a+b)=\lim_{x\to3^-}(ax^2+bx+2)$
implies
2. $\displaystyle 9-a+b=9a+3b+2$
From 1. we see that $\displaystyle a=-b$. So we substitute for $\displaystyle a$ in 2.
$\displaystyle 9-(-b)+b=9(-b)+3b+2$
We now solve for b
$\displaystyle 9+2b=-6b+2$
$\displaystyle 8b=-7$
$\displaystyle b=\frac{-7}{8}$.
And from 1.
$\displaystyle a=-\left(\frac{-7}{8}\right)$.
Therefore
$\displaystyle f(x)=\left\{\begin{array}{cc}\frac{x^2-1}{x-1},&\mbox{ if }
x<1\\\\{\frac{7}{8}x^2+\frac{7}{8}x+2},&\mbox{ if }1\leq{x}<3\\\\{3x-\frac{7}{4}},&\mbox{ if }x\geq3\end{array}\right.$