Question

Calculus! I need help once again! (continuity)?

Find the values of (a) and (b) that make (f) continuous everywhere (this is a piecewise function by the way):

f(x) =

(x^2-4)/(x-2) if x < 2

ax^2-bx+3 if 2<x<3

2x-a+b if x is greater than or equal to 3

Calculus is definitely not my subject. I absolutely hate it! Thanks in advance if you can help!

Answer

Here you want to make the function from the left of 2 meet the function from the right of 2 at the same y-value. That is, set the two equal to eachother:

(x^2-4)/(x-2) = ax2-bx+3

Factor the left side as a difference of squares to make things simpler:

(x+2)(x-2)/(x-2) = ax2-bx+3

(x+2) = ax2-bx+3

And set x to 2:

(2+2) = 4a - 2b + 3

1 = 4a - 2b

Now do the same at x=3:

ax2 - bx + 3 = 2x - a + b

9a - 3b + 3 = 6 -a + b

10a - 4b = 3

Multiply the first equation by -2 and solve by linear combination:

3 = 10a - 4b

-2 = -8a + 4b

------------------

1 = 2a + 0

a = 1/2

And solve for b!

1 = 4a - 2b

1= 2 - 2b

-1 = -2b

b=1/2

a=b=1/2

Unanswered