Question

Suppose $f\left(x\right)=$ $\{\begin{array}{cc}a+bx,& x<1\\ 4,& x=1\\ b-ax& x>1\end{array}$ and if $\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$, then what is the possible value of $a,b$?

Answer 1

We have,
$f\left(x\right)=\{\begin{array}{cc}a+bx,& x<1\\ 4,& x=1\\ b-ax& x>1\end{array}$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1}{lim}(a+bx)=a+b$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}(b-ax)=b-a$
Also $f\left(1\right)=4$
It is given that $\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)\cdot$

$\therefore$$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$

$\Rightarrow a+b=4$ and $b-a=4$
Solving these two equations we obtain $a=0$ and $b=4$