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

Question

f(x) =

(x² − 4)/(x − 2)             if x < 2
ax² − bx + 3                 if 2 ≤ x < 3
2x − a + b                    if x ≥ 3

Answer 1

For the function to be continuous at x = 2, the one sided limits at x = 2 must be equal.  

 

lim_x→2^- f(x)

= lim_x→2^- [(x²-4)/(x-2)]=lim_x→2^- [(x+2)(x-2)/(x-2)]=lim_x→2^- [x+2]=4 

 

 

lim_x→2⁺f(x) = lim_x→2⁺ (ax²-bx+3) = 4a-2b+3

 

So, 4a-2b+3 = 4.  Therefore, 4a-2b = 1

 

For f(x) to be continuous at x = 3, the one sided limits of f(x) at x = 3 must be equal.

 

lim_x→3^- f(x) = lim_x→3^- (ax²-bx+3) = 9a-3b+3

 

lim_x→3⁺ f(x) = lim_x→3⁺ (2x-a+b) = 6-a+b

 

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

 

Therefore, 10a -4b = 3

 

We have the system of equations:  4a - 2b = 1

                                                          10a - 4b = 3

 

Multiply the first equation by -2 and then add the equations:

 

-8a + 4b = -2

10a - 4b = 3

 

2a = 1    a = 1/2

 

4(1/2) - 2b = 1

 

2 - 2b = 1

 

   -2b = -1

 

      b = 1/2

 

So, if a = b = 1/2, then f(x) is continuous for all x.  

 

 

 

Answer 2

Continuous means that the graph of f(x) has no gaps in the intervals.

 

f(x) is a piecewise function because for each interval, there are different types of behaviors.  

 

 

Lets look at the first behavior.

 

(x² - 4) / (x - 2)

 

x cannot equal 2 due to the denominator being zero.   That is why x=2 is not included in that domain.  However, if we expand the numerator and cancel, it simplifies to (x + 2).  So you can say that f(x) is (x + 2) in the interval (-∞, 2).

 

We need to evaluate (x + 2) when x=1.99999.  We will use the f(x) value for the second piece.

 

(1.9999 + 2) = 3.9999

 

This means when we go to the second piece, which starts at x=2, the f(x) value would be 4.  We do this because the first piece needs to be connected to the second piece in order for it to be continuous.  Think of it like this:  as x gets closer to 2 from the left, f(x) gets closer to 4.  This is what we call a limit.

 

Now we go on to the second and third pieces.  We use the domains of these pieces to find a and b and the value of (x + 2) at x=1.9999 to create an equation.

 

 

Second piece:

 

a(2)² - b(2) + 3 = 4

 

4a - 2b + 3 = 4

 

4a - 2b = 1

 

 

Next, we can give the parabola a vertex.  If the x coordinate of the vertex lies between 2 and 3, then the endpoint of the parabola will be f(x) = 4.  This is because it is symmetrical.

 

Now to make the second piece and third piece share the same point, we equate them.   They will share the same point at x=3.

 

a(3)² - b(3) + 3 = 4

 

9a - 3b = 1

 

 

2(3) - a + b = 4

 

6 - a + b = 4

 

-a + b = -2

 

 

Now, we use these equations to solve for a and b.

 

9a - 3b = 1                 eq1

-a - b = -2                   eq2

 

 

Multiply eq2 by 9.  Keep eq1.

 

9a - 3b = 1               eq1

-9a - 9b = -18           eq2

 

 

Add both equation to eliminate the a terms.

 

-12b = -17

 

     b = 17 / 12

 

 

The substitute this value of b into either of these equations to solve for a.  Once you have a and b, plug them into the functions and test them out on a graphing calculator by plotting the three pieces to see if the graph is continuous.

 

 

I hope this helps.  Even I found this to be a super challenge.