Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{x^2-9}{x-3}+\alpha & ,\text{for}x>3\\ 5& ,\text{for}x=3\\ 2x^2+3x+\beta & ,\text{for}x<3\end{array}$

is continuous at $x=3$, find $\alpha$ and $\beta$.

Answer 1

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

For $x=3,f\left(x\right)=5$

For $x<3,f\left(x\right)=2x^2+3x+\beta$

Now, for continuity at $x=3$,

$f\left(3^{-}\right)=f\left(3\right)=f\left(3^{+}\right)$

$\therefore 2(3{)}^2+3(3)+\beta =5=3+3+\alpha$

$\therefore 27+\beta =5=6+\alpha$

$\therefore \alpha =-1\text{and}\beta =-22$

This is the required solution