Question

if  `f(x) = (x^2-9)/(x-3) + alpha`               for x> 3

           =5,                                     for x = 3

          `=2x^2+3x+beta`,             for x < 3

is continuous at x  = 3, find α and β.

Answer 1

∵f is continuous at x  = 3

∴ `lim_(x->3^(-)) f(x) = lim_(x->3^(+)) = f(3)`   ......(1)

Now `lim_(x->3^(+)) f(x) = lim_(x->0)((x^2-9)/(x-3) + alpha) = lim_(x->3)[((x-3)(x+3))/(x - 3) + alpha]`

`= lim_(x->0)[(x+3) + alpha] = (3+3) + alpha = alpha + 6`

and

`lim_(x->3^(-)) f(x) = lim_(x->3) (2x^2 + 3x +beta) = 2(3)^2 + 3(3) + beta = 18 + 9 + beta = 27 + beta`

Also f(3) = 5 ...(given)

∴ From (1) we get α + 6 = 27 + β = 5`

∴ α + 6 = 5 and 27 + β = 5

∴ α = 5 - 6 = -1 and β = 5 - 27 = - 22

⇒ ∴ α = -1 and β = -22