Find the values of a and b that make f continuous everywhere

Question

(x²-4)/(x-2) if x < 2

ax² - bx +1 if 2 <_ x < 3

4x-a-b if x >_ 3

Answer 1

For f(x) to be continuous when x = a, we must have lim_x→a^-f(x) = lim_x→a⁺f(x)

(x²-4) / (x-2) = (x+2)(x-2)/(x-2) = x+2

lim_x→2^-f(x) = lim_x→2^-(x+2) = 4

lim_x→2⁺f(x) = lim_x→2⁺(ax² - bx + 1) = 4a - 2b + 1

For f(x) to be continuous at x = 2, we must have 4a - 2b + 1 = 4. So, 4a - 2b = 3.

lim_x→3^-f(x) = lim_x→3^-(ax² - bx + 1) = 9a - 3b + 1

lim_x→3⁺f(x) = lim_x→3⁺(4x - a - b) = 12 - a - b

For f(x) to be continuous at x = 3, we need 9a - 3b + 1 = 12 - a - b. So, 10a - 2b = 11.

4a - 2b = 3

10a - 2b = 11

Subtract the equations to get -6a = -8. So, a = 4/3

4(4/3) - 2b = 3

-2b = -7/3

b = 7/6