Find the values of b and c that make f continuous everywhere.

Question

The function:

f(x)= (x²-6x+b)/(x²-7x+12) if x>4 and

2e^{x^2-3x+c} if x<=4

Thank you!

Answer 1

1. f(x) = (x² -6x +b)/(x2 - 7x + 12) ,x>4

= 2e ^{x2-3x +c} ,x≤4

You are required to find the values of b and c that make the piece wise function continuous at x = 4.

Rewrite, (x² -6x +b)/(x2 - 7x + 12) = (x² -6x +b)/(x -4)(x-3)

Strategy.

Can we find a value for b such that there is a hole at x = 4 for (x² -6x +b)/(x -4)(x-3)?

By inspection, if b = 8, then x² -6x + 8 = (x-4)(x-2) and

(x² -6x +b)/(x -4)(x-3) = (x-2)/(x-3)

The y value of the hole at x = 4 is (4-2)/(4-3) = 2

For the function to be continuous at x = 4, the y value of the hole must equal f(4).

i.e. f(4) = 2 →2e^{16-12 + c} = 2

At this point, I leave it to you to find the value of c - a simple exercise.