## Solution – Find the values of a and b that make the function differentiable everywhere.

Find all values of $$a$$ and $$b$$ that make the following function differentiable for all values of $$x$$.

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ ax+b & \mbox{if } x>-1 \end{cases}$$

When trying to solve a problem like this, there are actually two things you will need to consider for our function $$f(x)$$.  Obviously, we need to make sure that it’s differentiable everywhere, but this actually implies something else that we will want to consider as well.

Since a function being differentiable implies that it is also continuous, we also want to show that it is continuous.  The reason for this is that any function that is not continuous everywhere cannot be differentiable everywhere.  Once we make sure it’s continuous, then we can worry about whether it’s also differentiable.

## Making sure f(x) is continuous everywhere

I’m not going to go into quite as much detail to show the part about making sure the function is continuous because I have already done this, which you can see by clicking here.

To make sure $$f(x)$$ is continuous at $$x=-1$$ we need to make sure that $$\lim_{x \to -1} f(x) = f(-1).$$  Since we have a piecewise function, we will need to consider each one-sided limit, but in this case only the right sided limit will tell us something useful.

$$\lim_{x \to -1^{+}} f(x) = f(-1)$$

$$\lim_{x \to -1^{+}} ax+b = b(-1)^2-3$$

$$a(-1)+b=b-3$$

$$-a+b=b-3$$

$$-a=-3$$

$$a=3$$

So now we know that $$f(x)$$ will be continuous everywhere as long as $$a=3$$.  However, this doesn’t really tell us that $$f(x)$$ is differentiable everywhere as well.

## Making sure f(x) is differentiable everywhere

We now know that we will need to let $$a=3$$ in order for this function to be continuous and to have a chance of being differentiable.  As a result, we can say that we are now trying to make this function differentiable everywhere:

$$f(x) = \begin{cases} bx^2-3 & \mbox{if } x\leq -1 \\ 3x+b & \mbox{if } x>-1 \end{cases}$$

We can see that the only place this function would possibly not be differentiable would be at $$x=-1$$.  The reason for this is that each function that makes up this piecewise function is a polynomial and is therefore continuous and differentiable on its entire domain.  The only place we may have a problem is when we have to switch between the two functions.

#### What does it mean for a function to be differentiable?

It means that its derivative exists for all values of $$x$$.  In other words, we need to be able to find its derivative no matter what $$x$$ is.

However, as I mentioned above, in this case we really only need to make sure that we can find the derivative of $$f(x)$$ when $$x=-1$$ since we know it would exist for all other values of $$x$$.  By using the definition of a derivative, we need to make sure the following limit exists at $$x=-1$$.

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Since we need to check this when $$x=-1$$, we can plug in $$-1$$ for $$x$$.  Therefore, we need to make sure this limit exists:

$$\lim_{h \to 0} \frac{f(-1+h)-f(-1)}{h}$$

I went over this limit definition in greater detail previously.  If you want a refresher on where this is coming from you can find that by clicking here.

Just like when we had to find the limit to make sure that $$f(x)$$ was continuous, we will need to consider each one sided limit separately in order to find this limit.  And also like when we checked for continuity, each one sided limit is going to require the use of a different section of our piecewise function.

#### Setting up the limits

When $$h$$ is slightly less than $$0$$, and we are considering the left sided limit, $$f(-1+h)$$ would need to be found using the $$y=bx^2-3$$ because this would involve inputting $$x$$ values which are less than $$-1$$.

By the same reasoning, when $$h$$ is slightly greater than $$0$$, and we are considering the right sided limit, $$f(-1+h)$$ would need to be found using the $$y=3x+b$$ because this would involve inputting $$x$$ values which are greater than $$-1$$.  Therefore, we need to consider the following one sided limits:

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

#### Now what do we do with these limits?

Now remember, as I discussed in the lesson about one-sided limits, in order for a limit to exist we need both of its one-sided limits to exist and they need to be equal.  Therefore, in order to show that the derivative of $$f(x)$$ exists at $$x=-1$$, these two one-sided limits need to be equal to each other.  Before setting them equal to each other, first we’ll simplify them a bit.  First the left side limit.

$$\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)^{2}-3\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(-1+h)(-1+h)-3\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b(1-2h+h^2)-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{\Big[b-2bh+bh^2-3\Big]-\Big[b-3\Big]}{h}$$

$$=\lim_{h \to 0^{-}} \frac{b-2bh+bh^2-3-b+3}{h}$$

$$=\lim_{h \to 0^{-}} \frac{bh^2-2bh}{h}$$

$$=\lim_{h \to 0^{-}} \frac{h(bh-2b)}{h}$$

$$=\lim_{h \to 0^{-}} bh-2b$$

$$=-2b$$

And now the right sided limit.

$$=\lim_{h \to 0^{+}} \frac{\Big[3(-1+h)+b\Big]-\Big[b(-1)^{2}-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{\Big[-3+3h+b\Big]-\Big[b(1)-3\Big]}{h}$$

$$=\lim_{h \to 0^{+}} \frac{-3+3h+b-b+3}{h}$$

$$=\lim_{h \to 0^{+}} \frac{3h}{h}$$

$$=\lim_{h \to 0^{+}} 3$$

$$=3$$

Now if we set these two simplified versions of the one-sided limits equal to each other, we get

$$-2b=3$$

$$b=-\frac{3}{2}$$

#### What does this tell us?

So now if we put both pieces together, we know that $$a=3$$ will ensure that $$f(x)$$ is continuous and then making $$b=-\frac{3}{2}$$ will also make sure $$f(x)$$ is differentiable at $$x=-1$$.  This would in turn make $$f(x)$$ differentiable for all values of $$x$$, or make it differentiable everywhere.

As always, I want to hear your questions!  Go check out my other lessons about derivatives and if you can’t get your question answered, I’d love to hear from you.  Leave a comment below or email me at jakesmathlessons@gmail.com.  If you have questions on this problem and solution or if you have another question you would like to see me answer, just ask it.  Or if you have an entire topic you would like to see me write a lesson about, just let me know.

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## Solution – Find the values of a and b that make f continuous everywhere

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:

1. making each individual piece continuous on its reduced domain,
2. 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 $$f$$ is actually defined by that function.

Piece 1: $$y=\frac{x^2-4}{x-2}$$ when $$x<2$$

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 $$x=2$$.  And in fact, this is the only value for x where $$y=\frac{x^2-4}{x-2}$$ 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 $$x=2$$ is NOT in the domain $$x<2$$, 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: $$y=ax^2-bx+3$$ when $$2\leq x<3$$

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 $$x^2$$ 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 $$2\leq x<3$$.  Therefore, we don’t need to worry about our a and b causing any discontinuities here.

Piece 3: $$y=2x-a+b$$ when $$x\geq 3$$

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 $$x\geq 3$$.

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 $$x=2$$ (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 $$x=3$$.

Continuity at $$x=2$$

Remember by definition, a function f(x) is continuous at $$x=a$$ 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 $$x=2$$. 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 $$x=2$$ 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 $$x<2$$.  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 $$(x-2)$$ term so we can use them to cancel each other out.  This leaves us with

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

$$y=x+2$$ is a linear function and is continuous everywhere, so we can simply plug in $$x=2$$ 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 $$x=2$$ 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 $$x=2$$.  We know that $$f(x)=ax^2-bx+3$$ when $$2\leq x<3$$.  We can use this part to find f(2) because this is the piece that defines f when $$x=2$$.  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 $$x=2$$ 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 $$y=ax^2-bx+3$$ 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 $$x=2$$ as long as

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

At this point we have found a set of a and b values that make f continuous at $$x=2$$.  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 $$x=3$$.  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 $$x=3$$

Making sure that f is continuous at $$x=3$$ will be an extremely similar process to what we just did around $$x=2$$.  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 $$x=3$$ 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 $$x\geq 3$$.

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

Since a and b are both constants, $$y=2x-a+b$$ 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 $$x=3$$, 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 $$x=3$$ for x values slightly smaller than 3, we will want to use the piece of f that is defined for $$2\leq x<3$$.  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 $$y=ax^2-bx+3$$ is a polynomial, it is continuous everywhere and we can find this limit by plugging in $$x=3$$.

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

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

$$10a-4b=3$$

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 $$x=2$$ 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 $$x=2$$ and at $$x=3$$ if $$a=\frac{1}{2}$$ and $$b=\frac{1}{2}$$.  We already showed that f is continuous for $$x<2$$, $$2<x<3$$, and $$x>3$$ for all values of a and b.  Now we also know that f is continuous at $$x=2$$ and $$x=3$$ if $$a=\frac{1}{2}$$ and $$b=\frac{1}{2}$$.  Therefore, we can say that f would be continuous everywhere if $$a=\frac{1}{2}$$ and $$b=\frac{1}{2}$$.

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

Check out the other topics I’ve covered and the problems I’ve worked through.  You can see a list of related lessons on my limits page.

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## Finding Derivatives with Limits

At this point we should have at least a basic understanding of limits and how to find some limits.  However, I have only really discussed limits by themselves and not how they relate to the rest of calculus.  They are very important in calculus because they are used to define the most important calculus topics.

For example, the main topic which will be discussed for quite some time is derivatives.  Derivatives will come up in a lot of different settings, like finding rate of change, instantaneous rate of change, velocity, slope, and a few others.  The main thing to realize is that a derivative is generally used to find out how quickly, or slowly, something is changing.

I will go further into all of these things later, but for now I want to focus on the definition of derivatives and how to find a derivative using the definition.

## The definition of a derivative

If we have some function, $$f(x)$$, we would write “the derivative of f” as $$f'(x)$$.  And we would define the derivative of f by using this limit:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$

This limit can be a bit confusing, so there’s something I would like to point out before we actually begin working with this limit.  The confusing thing here is that we have $$x$$ and $$h$$ in this limit and it looks as if they are both variables.  However, when we find this limit, we can only treat $$h$$ as a variable.  We will need to treat x as a constant and h as the only variable.

The reason for this is that we are finding the limit as $$h$$ goes to $$0$$.  This tells us that $$h$$ is moving in toward $$0$$.  It does not tell us that $$x$$ is changing at all.  Therefore, when we are working with the limit, we will act as if $$x$$ is a number, or a constant.  This means that once we find the limit, our answer may have $$x$$ in it still and this is completely fine since $$x$$ isn’t the variable in this case.  Now let’s try an example.

## Example 1

Consider the function $$f(x) = 4x^2 – 7x + 12$$.  We will use the limit definition to find the derivative of this function, but first let’s break it down and consider each part on its own.

### Finding f(x+h)

The fist thing we need to find is $$f(x+h)$$.  This notation basically just means that we need to look at our function $$f$$, and plug in $$(x+h)$$ wherever we see the input.  In other words, we need to replace all of the $$x$$’s in the function with $$(x+h)$$’s.  So,

$$f(x+h) = 4(x+h)^2 – 7(x+h) + 12.$$

Then we will want to expand this out so it’s easier to work with.  Remember $$(x+h)^2$$ is the same as $$(x+h)(x+h)$$, which means we need to foil it.

$$f(x+h) = 4(x+h)(x+h) – 7(x+h) + 12$$

$$=4(x^2 + xh + xh + h^2) – 7(x+h) + 12$$

$$=4(x^2 + 2xh+ h^2) – 7(x+h) + 12$$

$$=4x^2 + 8xh+ 4h^2 – 7x – 7h + 12$$

Since there aren’t any like terms we will leave it at that for now.

### Putting it all together

Now we can put that into the rest of the equation.  Since we now know $$f(x+h)$$ and $$f(x)$$, we can plug those into the equation.  I would recommend surrounding each of them with a set of parenthesis so you don’t forget to distribute the negative sign in front of the $$f(x)$$.  This is a very common mistake, so be careful not to forget that because it will give you the wrong answer.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$

$$= \lim_{h \to 0} \frac{(4x^2 + 8xh+ 4h^2 – 7x – 7h + 12) – (4x^2 – 7x + 12)}{h}$$

### Solving the limit

When I first see a limit, the first thing I usually consider is whether we can simply plug in $$0$$ for $$h$$. Essentially, I try to treat this function as if it were continuous at $$h=0$$ (remember $$h$$ is the variable here).

However, if we do this here we will get $$0$$ on the denominator.  Since you cannot divide by 0, this will not work.  So our strategy will be to simplify this fraction to a point where we can plug in $$0$$ for $$h$$.  The simplest way to do this is to rearrange the numerator so that we can cancel an $$h$$ from the numerator and denominator and get rid of our fraction all together.

$$f'(x)= \lim_{h \to 0} \frac{4x^2 + 8xh+ 4h^2 – 7x – 7h + 12 – 4x^2 + 7x – 12}{h}$$

$$= \lim_{h \to 0} \frac{8xh+ 4h^2 – 7h}{h}$$

At this point I would like to point something out. Notice, after simplifying the numerator of the fraction, each term remaining contains an $$h$$ in it.  This is important because it allows us to factor the $$h$$ out and cancel it with the $$h$$ in the denominator, getting rid of the fraction.  This will be an extremely common strategy to use for finding the derivative of a function using the limit definition.

$$f'(x)= \lim_{h \to 0} \frac{h(8x+ 4h – 7)}{h}$$

$$= \lim_{h \to 0} 8x+ 4h – 7$$

Now we have simplified to a point that we can solve this limit by plugging $$0$$ in for $$h$$.

$$f'(x)= 8x+ 4(0) – 7$$

$$= 8x – 7$$

So we have just shown that if $$f(x)=4x^2-7x+12$$, then $$f'(x)=8x-7$$.  We will later learn shortcuts like the product rule, quotient rule, and chain rule that will make finding derivatives like this much simpler and faster, but you will need to know how to find these using the limit definition.  The pattern shown in this problem is a common one.  It won’t work for all derivatives, but it’s a good thing to try first.

• It’s generally a good idea to see if you can reengage the top of the fraction in such a way that every term has $$h$$ as a factor.
• Then you can factor out the $$h$$, and cancel it with the $$h$$ on the bottom of the fraction.
• This usually leaves you with a function that you can directly plug $$0$$ into $$h$$ and simplify from there, leaving you with a function that doesn’t contain any $$h$$’s, but usually contains $$x$$’s.

Enter your email below and I’ll send you my calculus 1 study guide which is packed full of helpful tricks and shortcuts to help you boost your scores in calc!

I would recommend checking out the other material I have about derivatives.  As I mentioned before, there are several shortcuts and methods that make this whole process a lot easier.  Go check out what I’ve written about on the derivatives page.  If you have a question that isn’t answered there just let me know by emailing me at jakesmathlessons@gmail.com.  I’ll do my best to answer any questions you send me and I may even post a lesson or full solution on it!

## Continuity

Continuity is a relatively simple concept, but problems that require proving it can be a little tricky. Essentially, a continuous function is one that you can draw all in one motion without picking up your pencil. This is one explanation of what it means for a function to be continuous that I like because it doesn’t take any mathematical definitions or proofs to understand. Any holes or gaps in a function’s graph would be a discontinuity and would mean that the function is not continuous.

## The limit definition of continuity

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

Let’s think about what this equation is saying. The left side of this equation is something that we’ve already dealt with, limits. It’s asking us to find out what $$y$$ value we close in on as we travel along our function and close in on $$x=a$$. Keep in mind, $$x$$ is a variable here which represents the input of our function, $$f(x)$$, and $$a$$ is a constant. This means that $$a$$ represents some specific number, which could be any number.

The right side of the equation is simply asking us to plug that same $$a$$ value into $$f(x)$$ and take the $$y$$ value we get out.

In total, the above equation says that if we travel along a function and close in on a specific $$x$$ value, we should close in on the same $$y$$ value we would get if we simply plugged that $$x$$ value into the equation.

In other words, as we travel along a function toward a specific $$x$$ value, our $$y$$ value will also go toward the $$y$$ value of the function at that point. If that is the case, then the function is continuous at that specific point. If we can say that a function is continuous at every single possible value we could put in for $$a$$, then we can say that the function is continuous for all $$x$$. If this is true then we can draw our entire function in one motion without picking up our pencil!

Let’s do a couple examples.

## Example 1

Remember back in the first lesson about limits, Limits – Intro, I said I would go back to discussing the importance of our limit in the first example giving us the same value as when we plugged $$x=2$$ into the equation?  I would like to go into that further.

The function we were considering was $$f(x) = x^2$$ and we were finding

$$\lim_{x \to 2} x^2.$$

After looking at the graph of this function, shown in Figure 1.1, we saw that

$$\lim_{x \to 2} x^2 = 4.$$

I also pointed out that plugging in $$x=2$$ directly into the function also returns a $$y$$ value of $$4$$. In other words, we know $$f(2) = 4.$$ Therefore, we know

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

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

Notice this is exactly like the definition of what it means for a function to be continuous at a point. If you replace $$a$$ with $$2$$, replace $$f(x)$$ with $$x ^2$$, and replace $$f(a)$$ with $$4$$, we have just shown that $$y=x^2$$ is a function that is continuous at $$x=2.$$

## Example 2

The next example I want to discuss also goes back to a function we have already looked at. Referring back to Limits – Intro we will consider $$f(x)$$ shown in Figure 1.2.

The question we will answer here is whether this function is continuous at $$x=1$$ or not. What do you think?

There are a few different ways we can answer this question. The simplest would be to simply look at the graph of the function and think about whether we can draw that function at and around $$x=1$$ without picking up our pencil. As you can see, there is clearly a hole at $$x=1$$ where we would need to pick up our pencil, and add a single point at $$(1, \ 4)$$.

As a result, it is probably safe to say that this function is not continuous at $$x=1$$. However, we want to be able to show this using the actual definition of what it means for a function to be continuous.

Remember, for this function, which we are calling $$f(x)$$ in this case, we need to be able to show that $$\lim_{x \to 1} f(x) = f(1).$$

If we can show this equation to be true, then $$f(x)$$ is continuous at $$x=1$$, and if it’s not true then the function is not continuous at $$x=1$$. Luckily, back when I first used this function as an example in the Limits – Intro lesson, we found that $$\lim_{x \to 1} f(x) = 2.$$ Therefore, we just need to find out if $$f(1)$$ is also $$2$$ and we can prove that this function is continuous or not continuous.

Looking at the graph again, we see that this function has a hole at $$x=1$$ and includes the point $$(1, \ 4)$$. In other words, if we plug $$1$$ into $$f(x)$$ as our $$x$$ value, we get a $$y$$ value of $$4$$ out. This is the same as saying $$f(1) = 4$$.

Now, at this point we have figured out $$\lim_{x \to 1} f(x) = 2 \ and$$ $$f(1) = 4.$$

Therefore, $$\lim_{x \to 1} f(x) \neq f(1)$$ and we can say that $$f(x)$$ is not continuous at $$x=1$$.

## More Examples

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

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

I also have a calc 1 study guide that you can get now for FREE! Just enter your name and email below and I’ll send you a copy straight to your inbox!

There are also several other lessons and problems on my limits page.  It would be a good idea to get some practice with limits.  A lot of other more complex topics in calculus are based around limits so they are important to understand.  If you can’t find the topic you want to read about just let me know by emailing me at jakesmathlessons@gmail.com and I’ll do my best to answer your question!