Answer:
Step-by-step explanation:
For a function to be continuous
The right hand limit f'+(x) must be equal to the left hand limit f'-(x)
The right hand limit exist at where x>3
The resulting function is. x²+a
If x = 3
f'+(3) = 3²+a
f'+(3) = 9+a
For the left hand limit, this exist where
bx+a
f'-(x) =bx+a
f'-(3) = 3b+a
Since f'+(3) = f'-(3)
9+a = 3b+a
9 =3b
b = 9/3
b = 3
At the end point x = -3
For the left hand limit
f-(x)= √-b-x
f-(-3) = √-b-(-3)
f(-3) = √-b+3
For the right hand, the function is bx+a
f'+(x) = bx+a
f+(-3) = -3b + a
Equate both limits
√-b+3 = -3b+a
√-3+3 = -3(3)+a
0 = -9+a
a= 9
Hence a = 9 and b = 3
