If `f(x)(x^(2)-9)/(x-3)+alpha, "for" x gt 3`
`=5, "for" x =3`
`=2x^(2)+3x+beta, "for" x lt 3`
is continous at `x=3, ` find `alpha and beta. `

Explanation

`:' ` f (x) is continuous at x = 3, we get
`underset(x to 3^(+))lim f(x) = underset(x to 3^(-))lim f(x) = f(3)`
`f(3) = 5` (Given)
` :. " " underset(x to 3^(+)) lim f(x) = f(3)`
` rArr " " underset (x to 3) lim (x^(2) - 9)/(x - 3) + alpha = 5`
` underset(x to 3) lim ((x+3)(x-3))/((x-3)) + alpha = 5`
` rArr" " underset( x to 3) lim x + 3 + alpha = 5`
` rArr 3 + 3 + alpha = 5`
` rArr " " 6 + alpha = 5`
` rArr " " alpha = 5 - 6 = - 1`
Also `underset(x to 3^(-)) lim f(x) = f (3)`
` rArr " " underset(x to 3) lim 2x^(2) + 3x + beta = 5`
` rArr" " 2 * 3^(2) + 3*3+ beta = 5`
` 18 + 9 + beta = 5`
` rArr " " 27 + beta = 5`
` rArr" " beta = 5 - 27 = - 22`
` rArr" " alpha = - 1, beta = - 22 `