Question

B. Determine the values of a and b so that f is continuous

f(x)= { x^2 +a if x > or = 3

bx+a if x is > 3 and if x is less than 3

root -b-x if x < or = -3

Answer 1

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