Consider the following piecewise function: $$f(x) = \begin{cases} \frac{x^2-4}{x-2} & \mbox{if } x<2 \\ ax^2-bx+3 & \mbox{if } 2\leq x<3 \\ 2x-a+b & \mbox{if } x\geq 3 \end{cases}$$

Problem credit: Chapter 2.4 #36 in Single Variable Calculus: Concepts and Contexts by James Stewart.

## How to find *a* and *b* for a piecewise function to be continuous everywhere.

When we see piecewise functions like this and our goal is to make sure it is continuous everywhere, there are two cases we want to consider separately:

- making each individual piece continuous on its reduced domain,
- and making the different pieces line up when our function switches between them.

First let’s consider each piece individually on the reduced domain where is actually defined by that function.

* Piece 1:* when

When considering a function to find *x* values where the function is not continuous, two very common things to look for would be taking the square root of a negative number or dividing by zero. Obviously we don’t need to worry about taking the square root of a negative number here since we don’t have any square roots. However, we do have a fraction, and therefore we should consider if there is an *x* value that would cause us to divide by zero.

Looking at the denominator of our fraction, we see that we would divide by zero when we plug in some number for *x* that causes

$$x-2=0.$$

Clearly, this would happen when . And in fact, this is the only value for *x* where is not continuous. However, remember that I said we want to see if there are any *x* values that cause the function not to be continuous **within** our limited domain for the specific piece of the function. Since is **NOT** in the domain , we can say that this piece of our function is continuous on its entire domain and won’t contribute any discontinuities to *f*.

* Piece 2*: when

Looking at this piece of our piecewise function, clearly we need to consider our constants *a* and *b.* Since our function *f* is a function of *x* (indicated by *f(x)*), we can consider the other letters in this piece of our function (*a* and *b*) to be constants. I discussed this in a bit more detail here, but it basically means that *a* and *b* are some set number, they do not change.

The important thing to realize though, is that regardless of what values we choose for *a* and *b*, this piece of our function will just be a polynomial. More specifically, it will be a quadratic function. No matter what our *a* and *b* end up being we will have an term, an *x* term, and a constant term. Since we know this piece of our function will be a polynomial no matter what our *a* and *b* are we can say that this piece of our function will be continuous for any *a* and *b* we select.

This is the case because polynomials are always continuous everywhere. If this section of our function would be continuous everywhere, it would certainly be continuous on our limited domain of . Therefore, we don’t need to worry about our *a* and *b* causing any discontinuities here.

* Piece 3*: when

This is actually going to be similar to *piece 2* above. We can see by looking at this function that no matter what *a* and *b* values we select here, they will both be constants which will combine to a single constant when we combine like terms. Once we do this, we will be left with an *x* term and a constant term. Therefore, this is also a polynomial (in this case it would also be a linear function), and would be continuous everywhere as a result. Because of this, it must be continuous on our limited domain of .

Now we have shown that we will not get any discontinuities **within** any of our restricted domains that make up our piecewise function, but what about on their edges. We also have to make sure we are selecting an *a* and a *b* value so that we can switch from one part of our function to the other without jumping up or down and leaving a hole in our function. If we do not select the correct *a* and *b* we may get discontinuities at the *x* values where we switch from one function to the other.

In order to prevent this, we need to make sure that our first two pieces of *f(x)* transition smoothly when (this is the place we switch between these two functions) and we need to make sure that the second and third piece of our function have a smooth transition between them at .

**Continuity at**

Remember by definition, a function *f(x)* is continuous at if $$\lim_{x \to a} f(x) = f(a).$$

We need to apply this rule here to make sure that our function *f* is continuous at . Therefore, we need to pick an *a* and a *b* so that $$\lim_{x \to 2} f(x) = f(2).$$

As I explained here, a limit will only exist if both of its one-sided limits exist and are equal to each other. Therefore, rather than finding the above limit, we actually want to find these two one-sided limits:

$$\lim_{x \to 2^{-}} f(x) = f(2)$$

$$\lim_{x \to 2^{+}} f(x) = f(2).$$

Let’s start with the left-sided limit. Since we are approaching from the left, we are looking at *x* values that are slightly less than *2*. Therefore, we need to use the piece of our function that is defined for . So the limit we really want to find, and the equation we want to solve is:

$$\lim_{x \to 2^{-}} \frac{x^2-4}{x-2} = f(2).$$

When attempting to solve this limit, we see that we can not just plug in *2* for *x* because this would cause us to divide by zero. As a result of this, I would first suggest simplifying the function whose limit we are trying to find. Using the *difference of squares* rule on the top of our fraction we can rewrite this limit as

$$\lim_{x \to 2^{-}} \frac{(x+2)(x-2)}{x-2} = f(2).$$

Once we do this, we can see that the top and bottom of our fraction contain a term so we can use them to cancel each other out. This leaves us with

$$\lim_{x \to 2^{-}} (x+2) = f(2).$$

is a linear function and is continuous everywhere, so we can simply plug in to find this limit.

$$4=f(2)$$

Now we need to figure out the right side of the equation. To find *f(2)*, we just need to plug in into our function *f(x)*, which is our original piecewise function. The first thing we need to do is decide which piece defines our function when . We know that when . We can use this part to find *f(2)* because this is the piece that defines *f* when . Plugging in *2* for *x* we get

$$4=ax^2-bx+3$$

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

$$4=4a-2b+3$$

$$1=4a-2b$$

Notice we have only one equation at this point which relates two unknown constants, making it impossible to find one unique solution for *a* and *b*. To do this we will need another equation relating these two constants.

I would also like to point out that this equation came from using the equation involving only the left sided limit, and we still need to look at the right sided limit.

$$\lim_{x \to 2^{+}} f(x) = f(2)$$

If we look at this limit, we see that we would be approaching from the right side, which would mean we are looking at *x* values slightly larger than *2*. For *x* values slightly larger than *2*, but infinitely close to *2*, we would use the piece to define *f*. We already found *f(2)* above, so putting these two facts together we see

$$\lim_{x \to 2^{+}} ax^2-bx+3 = a(2)^2-b(2)+3$$

Since the function whose limit we are trying to find is continuous everywhere (since it’s a polynomial) we can just plug in *2* for *x* to find this limit.

$$a(2)^2-b(2)+3=a(2)^2-b(2)+3$$

$$4a-2b+3=4a-2b+3$$

Notice both sides of this equation are the same. Because of this, this equation will actually be true no matter what we put in for *a* and *b*. This doesn’t really help us at all in this case, but it was important to test it out and see what it told us. So we know that *f* will be continuous at as long as

$$1=4a-2b.$$

At this point we have found a set of *a* and *b* values that make *f* continuous at . But we need to find just one *a* and one b that will accomplish this for all *x* values.

But this does not tell us anything about whether it would also be continuous at . Checking that may give us another relationship between *a* and *b* that we can use to find the single unique solution for the two constants that will make *f* continuous everywhere.

**Continuity at**

Making sure that *f* is continuous at will be an extremely similar process to what we just did around . Similar to above, we will need to make sure the following equations are true:

$$\lim_{x \to 3^{-}} f(x) = f(3)$$

$$\lim_{x \to 3^{+}} f(x) = f(3).$$

Let’s start with the right sided limit. This means that we are getting closer and closer to and we are coming from *x* values that are slightly larger than *3*. Therefore, we will need to use the piece of our function that defines *f* for .

$$\lim_{x \to 3^{+}} 2x-a+b = f(3)$$

Since *a* and *b* are both constants, is a linear function, and is continuous everywhere as a result. Because of this, we can just plug *3* in for *x* to find this limit.

$$2(3)-a+b = f(3)$$

$$6-a+b = f(3)$$

To find *f(3)* we just need to plug *3* in for *x* into the piece of our function that defines it when , which is the third piece of *f*.

$$6-a+b = 2(3)-a+b$$

$$6-a+b = 6-a+b$$

We see here another case where this equation gives us no useful information. This is because this equality will be true no matter what we plug in for *a* and *b*. Let’s move onto the left sided limit and see what we get there.

$$\lim_{x \to 3^{-}} f(x) = f(3)$$

Now that our *x* value is approaching *3* from the left side, it is slightly smaller than *3* and slowly increasing. Since we are considering what our function is doing around for *x* values slightly smaller than* 3*, we will want to use the piece of *f* that is defined for . We already found *f(3)* above, so we will also plug that in.

$$\lim_{x \to 3^{-}} ax^2-bx+3 = 6-a+b$$

Just like before, since we know is a polynomial, it is continuous everywhere and we can find this limit by plugging in .

$$a(3)^2-b(3)+3 = 6-a+b$$

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

$$10a-4b=3$$

**Putting it all together**

Now we have another relationship that relates *a* and *b*. Alone, it isn’t very useful, but we can take the previous equation we found from ensuring *f* is continuous at and solve the following system of equations:

$$\mbox{(1): }4a-2b=1$$

$$\mbox{(2): }10a-4b=3.$$

Now that we have two equations that relate these two unknown constants, we have enough information to solve for them. You always need at least as many equations as you have variables (or unknown constants in this case). There are a couple different ways to solve a system of equations like this, but I will use substitution. First we will solve for *b* in equation (1), then plug that into equation (2).

$$4a-2b=1$$

$$-2b=-4a+1$$

$$\mbox{(3): }b=2a-\frac{1}{2}$$

Now we will plug this into equation (2) and solve for *a*.

$$10a-4\Big(2a-\frac{1}{2}\Big)=3$$

$$10a-8a+2=3$$

$$2a+2=3$$

$$2a=1$$

$$a=\frac{1}{2}$$

Now we can plug this back into our equation for *b*, equation (3).

$$b=2\Big(\frac{1}{2}\Big)-\frac{1}{2}$$

$$b=1-\frac{1}{2}$$

$$b=\frac{1}{2}$$

Now we have shown that *f(x)* will be continuous at and at if and . We already showed that *f* is continuous for , , and for all values of *a* and *b*. Now we also know that *f* is continuous at and if and . Therefore, we can say that *f* would be continuous **everywhere** if and .

In fact, we can see that *f* is continuous by plugging in the values we found for *a* and *b* and graphing it using Desmos.

This type of problem is one that you will likely run into a few times. It actually comes in handy when you are trying to solve a similar type of problem that requires you to make a function differentiable everywhere. If you would like to see this I have worked through one of these and you can see that by clicking this link: Solution – Find the values of a and b that make the function differentiable everywhere.

## SHORTCUT – More examples

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