Question

f(x) = 1 + 3bx + 2x^2 x =< 1
f(x) = mx+b x > 1

Find the values of m and b so that f will be both continuous and differentiable at x=1

I got m = 1 and b = -1, but my work wasn't pretty and I'm not sure I believe my answer.

Answer 1

For f to be continuous at x = 1, we must have the two parts of the function equal, so

1 + 3bx + 2x^2 = mx + b at x = 1:

1 + 3b + 2 = m + b

3b + 3 = m + b

2b = m - 3. first equation [1]

For the function to be differentiable, it means the derivatives of the two parts must match. So:

f '(x) = {3b + 4x or m}

So we have

3b + 4x = m, at x = 1;

3b + 4 = m. second equation [2].

Solve [1] and [2] by substitution or solve for b or m and sub then:

[1] 2b = m - 3
[2] m = 3b + 4

so

2b = (3b + 4) - 3

2b = 3b + 1

-b = 1

b = -1, and with

[2] m = 3b + 4

= 3*(-1) + 4

= 1. So the answer is (m, b) = (1, -1)