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!

Answers

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

#1

you state that x<2 or 2<x so x is not 2, ever. this means it is discontinuous at 2.

#2

lim (x --> 2-) (x^2-4)/(x-2) = 4, so 4a - 2b +3 = 4, or 4a - 2b = 1 is required.

At x = 3, it is required that 9a - 3b + 3 = -a + b + 6, or 10a - 4b =3

(4 -2 1)

(10 -4 3)

(8 -4 2)

(10 -4 3)

2a = 1, a = 1/2

2 - 2b = 1

2b = 1

b = 1/2

so a = b = 1/2

#3

For this function to be continuous, the left- and right-hand limits at x=2 need to be the same; likewise when x=3. Set the first 2 equations equal to each other and solve for a when x=2; this will make the function continuous at x=2. Set the middle and last equations equal to each other and solve for a when x=3.

#4