Find the values of a and b so f has a derivative at x=2, namely that f’(2) exists.

Question

f(x)= {3x²-1 if x≤2

{ax+b if x>2

Answer 1

If the derivative exists then f(2) for 3x² - 1 must be the same value as f(2) for ax + b (the function must be continuous) AND, f '(2) must be the same for the two parts as well (no sharp points).

f '(x) = 6x for x ≤ 2 and a for x > 2

So, (plugging in x = 2) we can say:

1) 3(2)² - 1 = a(2) + b

and

2) 6(2) = a or a = 12

Plugging in a = 12 the the first equation, we get:

3(2)² - 1 = (12)(2) + b

12 - 1 = 24 + b

b = -13