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

Question

f(x)=x^2-4/x-2 if x<2

f(x)=ax^2-bx+3 if 2<x<3

f(x)=2x-a+b if x>3

Answer 1

The function is clearly continuous for x < 2, 2 < x < 3 and x > 3.

The only possibility of discontinuity is at x = 2 and x = 3.

By definition of continuity we need to make sure that

lim (x->2) f(x) = f(2) = 4a - 2b + 3 (from definition of f(x) when 2 <= x < 3)

lim (x->3) f(x) = f(3) = 6 - a + b (from definition of f(x) when x >= 3)

The first constraint involves limit f(x) when x < 2:

2² - 4/2 -2 = 4a - 2b +2

The second constraint involves limit of f(x) when 2 <= x < 3:

a3² - 3b + 3 = 6 - a + b

Thus we get two equations with two unknowns a, b:

4a -2b + 3 = 0

10a - 4b - 3 = 0

Solving them gives a = 4.5, b = 10.5

Answer 2

I answer to the same exercise here

https://www.wyzant.com/resources/answers/821676/find-the-values-of-a-and-b-that-make-f-continuous-everywhere

but the last equation was f(x)=4x-a+b instead of f(x)=2x-a+b.

Best,

Davide