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

Question

f(x)=x^2−4/x−2, if x >2

ax2−bx+1, if 2 ≤ x ≤3

4x−a+b, if x≥3

A = ?

B = ?

Answer 1

By setting the pairs of rules = to each other at the breaks in the domain (x = 2 and x = 3) we will get two equations in a and b that should then be solvable:

(I will assume that the top rule is meant to be for x < 2 NOT x > 2 as written, which wouldn't make sense.)

While the top expression is undefined at x = 2, the lim_x→2 (x²-4)/(x-2) = 4, which we can determine most easily by factoring and canceling the (x-2) from numerator and denominator.

For x = 2, the middle expression becomes 4a - 2b + 1, which we set = 4 from above. So we have 4a - 2b = 3.

For x = 3, the middle expression instead is 9a - 3b + 1, while the bottom is 12 - a + b, which expressions we set = to get 10a - 4b = 11. Multiplying the equation in a and b from x = 2 above we get 8a - 4b = 6. and subtracting that from the above equation eliminates b to give 2a = 5, a = 5/2 and b = 7/2.

This generates a point at (2,4) which makes the top and middle rules agree at x = 2, and a 2nd point at (3,13) which makes the middle and bottom rules agree at x = 3.