
Zeeshan I. answered 04/12/19
Differential Equations teacher for 2 years
The given function is f(x) = {x + 3 if x < 3; ax2 + bx if x ≥ 3.
For this function to be continuous, we must have that
limit x→3- f(x) = limit x→3+ f(x)
(3) + 3 = a(3)2 + b(3)
6 = 9a + 3b
or
4 = 6a + 2b ---(1)
The derivative of this function is f'(x) = {1 if x < 3; 2ax + b if x ≥ 3.
To be differentiable, we must have that
limit x→3- f'(x) = limit x→3+ f'(x)
1 = 2a(3) + b
1 = 6a + b ---(2)
Subtract equation (2) from equation (1) to get
4 - 1 = 6a - 6a + 2b - b
b = 3