find the value of the constant c that makes the following function continuous on (−∞ ∞)

Question

f(b) = cb + 3, bE(-INF, 4]

= cb^2 - 3, bE(4, INF)

Answer 1

Since f(b) = cb + 3 is continuous for any value of "c" and since f(b) = cb² - 3 is also continuous for any value of "c" then the only point we need to be worried about is what happens at b = 4 (the boundary between the two pieces).

The function value at b = 4 would be governed by f(b) = cb + 3 so f(4) = c•4 + 3 or f(4) = 4c + 3.

On the other interval, (4, ∞), we would consider the limit as b → 4 from the right. The value of that would equal cb² - 3 where b = 4 or "16c - 3"

To be continuous, the two parts must have the same value so:

4c + 3.MUST be equal to 16c - 3

4c + 3.= 16c - 3

6 = 12c or c = 1/2