Question

If $f\left(x\right)=\frac{x^2-9}{x-3}+\alpha ,\text{for}x>3$
= $5$ , for $x=$ $3$
= $2x^2+3x+\beta$, for $x<$ $3$
is continuous at $x=$ $3$, find $\alpha$ and $\beta$.

Answer 1

$f\left(x\right)=\frac{x^2-9}{x-3}+\alpha ,x>3$

$=5,x=3$

$=2x^2+3x+\beta ,x<3$

And $f\left(x\right)$ is continuous at $x=3$

For continuity, LHL and RHL at $x=3$ must be equal to $f\left(3\right)$

i.e., $\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}f\left(x\right)=f\left(3\right)$

$\to \underset{h\to 0}{lim}f(3-h)=\underset{h\to 0}{lim}f(3+h)=f\left(3\right)$

$\Rightarrow \underset{h\to 0}{lim}3-h+9+\alpha =\underset{h\to 0}{lim}[2{(3+h)}^2+3(3+h)+\beta ]=5[\because 3-h<3\&3+h>3]$

$\Rightarrow 3+9+\alpha =2\left(9\right)+9+\beta =5\to 2$

From equation $2,12+\alpha =5$

$\alpha =-7$

& $27+\beta =5$

$\beta =-22$