What are the values of a and b that makes piece-wise function f continuous and differentiable?

Question

Find the values of a and b that makes piece-wise function f continuous and differentiable.

a) Given

f(x) = {bx2 +x, for x ≤ 4

ax + 6, for x > 4}

b) Given

f(x) = {h(x); for x < 5

g(x); for x ≥ 5}

where

h(x) = ax2 + bx + 20

g'(x) = −17

f(5) = 10

Answer 1

Part A:

For the function to be continuous, the left limit = right limit = function value where the function changes at an x-value of 4...so set the equations equal to each other, and then substitute 4 for x.

Thus:

bx² + x = ax + 6

b(4)² + 4 = a(4) + 6

16b + 4 = 4a + 6

Just moving stuff around:

4a - 16b = -2 Call this Equation 1

For the function to be differentiable find and set the derivatives (using the power rule) of the pieces equal to each other and substitute 4 for x.

2bx + 1 = a

2b(4) + 1 = a

8b + 1 = a. Call this Equation 2.

2 equations, 2 variables gives you a system. Substitute 8b + 1 into Equation 1 for a:

4(8b + 1) - 16b = -2

32b + 4 -16b = -2

16b + 4 = -2

16b = -6

b = -6/16 = -3/8 = -.375

Then plug the value of b back into Equation 2 to find the value for a.

8(-3/8) + 1 = a

-3 + 1 = a

-2 = a

So a = -2, b = -3/8

Those are the values of a and b which make the function both continuous and differentiable.

If you need help with b, please schedule a one-on-one tutoring sessions with me and I would be happy to help you with Calculus