Question

In the following determine the value of ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ so that the given function is continuous:
$f\left(x\right)=\{\begin{array}{c}4,ifx\le -1\\ a{x}^2+b,if-1<x<0\\ \cos x,ifx\ge 0\end{array}$.

Answer 1

Given,

$f\left(x\right)=\{\begin{array}{cc}4,& \text{if}x\le -1\\ a{x}^2+b,& \text{if}-1<x<0\\ \cos x,& \text{if}x\ge 0\end{array}$

If $f\left(x\right)$ is continuous at $x=-1$, then

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

$\underset{x\to -1^{-}}{lim}4=\underset{x\to -1^{+}}{lim}a{x}^2+b$

$\Rightarrow a+b=4.....\left(1\right)$

Also, if $f\left(x\right)$ is continuous at $x=0$, then

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)$

$\underset{x\to 0^{-}}{lim}a{x}^2+b=\underset{x\to 0^{+}}{lim}\cos x$

$\Rightarrow b=\cos \left(0\right)=1$

Substituting the value of $b$ in equation $\left(1\right)$, we get

$a+1=4\Rightarrow a=4-1=3$

Hence $a=3$ and $b=1$.