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Continuity

                       

There are a lot of strange functions out there. In order to do meaningful analysis, we have to restrict our attention to functions that are reasonably well-behaved.

One way a function can be well-behaved is for it to be continuous. Visually, this means that its graph doesn’t have any “breaks” in it. In the graphs below, the function above is continuous, while the function below is not.

Actually, the notion of continuity will be applied at a point in the domain of the function. So, to be more precise: the function above is continuous at all real numbers \(x,\) while the function below is discontinuous at \(x = 2,\) but continuous at all the other \(x\) values.

We’d like a more rigorous definition of continuity at a point, because “having no breaks in the graph” is too vague. Fortunately, limits will help us!

Think carefully about each of the following proposed definitions. Think about what each means visually. Which one should we use for continuity? Which concept perfectly captures this intuitive idea of having no breaks?

A function \(f\) is continuous at a point \(a\) in its domain if:

We have our definition: A function \(f\) is continuous at a point \(a\) if \(\lim\limits_{x \to a}f(x)\) exists, and equals \(f(a).\) Again, note that continuity is defined for each point. Functions can be continuous at some points, and discontinuous at others. We sometimes say a function is continuous (without referencing a point) if it’s continuous at each point in its domain (i.e. at each point where it's defined).

Here’s sort of a trick question. Is the function \(f(x) = \frac{1}{x}\) continuous?

Here’s some further intuition about continuous functions that will be helpful. To say \(f\) is continuous at \(a\) means that when we look at points close to \(a\) and apply \(f\) to them, we get points close to \(f(a).\) To use a slightly different notation, if \(dx\) represents a very small amount, then when \(f\) is continuous at \(a,\) \(f(a + dx)\) is very close to \(f(a).\)

In this picture, for example, the function is continuous at \(a,\) so small changes to the input at \(a\) yield small changes to the output. But it’s discontinuous at \(b\), and you can see that a very small change to the input at \(b\) can yield a large change to the output.

We now have three ways of thinking about \(f\) being continuous at \(a:\)

  • (Definition) \(\lim\limits_{x \to a}f(x)\) exists, and equals \(f(a).\)
  • (Visually) The graph of \(f\) doesn’t have a break at \(a.\)
  • (Intuitively) Small changes to \(a\) yield small changes in \(f\)’s output.

What are examples of continuous functions? Almost everything you’ve ever considered! We won’t prove this here, but polynomials, trig functions, and exponentials and logarithms are continuous at every point of their domain. Furthermore, combinations of continuous functions are continuous: If \(f\) and \(g\) are continuous, then so are \(f+g,\) \(f \cdot g,\) \(\frac{f}{g},\) and \(f \circ g,\) at every point \(a\) on which they are defined.

How many of the following functions are continuous at every point in their domain?

  • \(\displaystyle 3x^2 + 2x - 5\)

  • \(\displaystyle \frac{1}{x-3}\)

  • \(\displaystyle \sin\left(\frac{1}{x}\right)\)

  • \(\displaystyle \Large e^{\cos\left(e^{\cos\left(e^{x^2}\right)}\right)}\)

It’s worthwhile to also consider piecewise-defined functions. How many of the following functions are continuous at every point in their domain? (It might help to graph each function.)

\[f(x) = \begin{cases}\begin{array}{rl} x^2, & x < 2 \\ x+2, & x \geq 2 \\ \end{array}\end{cases}\]

\[g(x) = \begin{cases}\begin{array}{rl} x^2, & x < 2 \\ x^2 + 1, & x \geq 2 \\ \end{array}\end{cases}\]

\[h(x) = \begin{cases}\begin{array}{rl} x^2, & x < 2 \\ (x-4)^2, & x \geq 2 \\ \end{array}\end{cases}\]

\[i(x) = \begin{cases}\begin{array}{rl} x^2, & x < 2 \\ 0, & x=2 \\ 4, & x > 2 \\ \end{array}\end{cases}\]

For one more example, let’s return to the function from the end of Quiz 2.

\[f(x) = \begin{cases}\begin{array}{rl} x, & x = \frac{1}{n} \text{ for some integer } n \\ 0, & \text{otherwise.} \\ \end{array}\end{cases}\]

At what points is this function continuous?

Hint: Don’t use the visual interpretation of continuity. Instead, use the definition. At what points \(a\) does \(\lim\limits_{x \to a}f(x)\) exist and equal \(f(a)\)?

A) At all points.

B) At all points except for those of the form \(\frac{1}{n}.\)

C) At all points except for 0 and those of the form \(\frac{1}{n}.\)

D) At all points except 0.

We started this quiz by saying that continuous functions are reasonably well-behaved. This good behavior is reflected in two basic theorems: the EVT (Extreme Value Theorem) and the IVT (Intermediate Value Theorem).

Suppose \(f\) is continuous at every point in an interval \([a,b].\) The Extreme Value Theorem says that \(f\) takes on a maximum value and a minimum value. In other words, there is at least one point \(p\) in \([a,b]\) such that \(f(p) \geq f(x)\) for all \(x\) in \([a,b],\) and there is at least one point \(q\) in \([a,b]\) such that \(f(q) \leq f(x)\) for all \(x\) in \([a,b].\)

For example, look at these two functions. Where is the maximum point guaranteed by the Extreme Value Theorem? It should be a point on the \(x\)-axis, in the interval \([1,3].\)

A) 5 on the left; 5 on the right.

B) 5 on the left; 3 on the right.

C) 2 on the left; 3 on the right.

D) 2 on the left; 2 on the right.

While the Extreme Value Theorem says something about maximum (and minimum) values, the Intermediate Value Theorem says something about the values in between. It says that if \(f\) is continuous on an interval \([a,b],\) then for every number \(z\) between \(f(a)\) and \(f(b),\) there is some \(p\) in \([a,b]\) for which \(f(p) = z.\) In other words, “all the intermediate values will get hit by \(f\).”

For instance, in the picture below, the IVT guarantees that there is some \(p\) between \(1\) and \(5\) for which \(f(p) = z.\) What is that \(p\)?

While the EVT and the IVT may seem “obvious”, it’s important to realize that their conclusions might fail if the function isn’t continuous! Which of the following discontinuous functions provides a counterexample to both the EVT and the IVT over the interval \([1,3]\)?

Note: We want an example where the EVT will fail. So there should be no \(p\) in \([1,3]\) which is greater than or equal to all the other function values. (You can assume it’s the “maximum” part, not the “minimum” part, that will fail.) And we want the IVT to fail. So there should be some \(z\) between \(f(1)\) and \(f(3)\) which is not hit by any point \(x\) in the interval.

A) \[f(x) = \begin{cases}\begin{array}{rl} x^2, & 1 \leq x < 2 \\ 5, & 2 \leq x \leq 3 \\ \end{array}\end{cases}\]

B) \[g(x) = \begin{cases}\begin{array}{rl} x^2, & 1 \leq x < 3 \\ 5, & x=3 \\ \end{array}\end{cases}\]

C \[f(x) = \begin{cases}\begin{array}{rl} x^2, & 1 \leq x < 2 \\ 0, & 2 \leq x \leq 3 \\ \end{array}\end{cases}\]

Since polynomials are continuous functions, one of these two theorems (EVT or IVT) is very useful for estimating where the zeros of polynomials are. Which theorem, and why?

A) The IVT: If \(p\) is a polynomial and we find points \(a\) and \(b\) where \(p(a)\) is negative and \(p(b)\) is positive, then by IVT \(p\) must have a zero somewhere in \([a,b].\)

B) The EVT: If \(p\) is a polynomial and we know that \(p(x) \geq 0\) for every \(x\) in some interval \([a,b],\) then by EVT there is some point \(q\) at which the polynomial takes on a minimum value, and therefore the polynomial must be 0 at \(q.\)

Here is how these theorems relate to things to come:

The Intermediate Value Theorem has a more sophisticated cousin, the Mean Value Theorem, which we’ll look at in the next chapter; it relates average and instantaneous rates of change.

The Extreme Value Theorem tell us maximum and minimum points exist, but not how to find them. Later in this course we’ll use calculus to find where the extreme values occur. This will allow us to solve all kinds of interesting optimization problems, like how to design a can with the largest volume from a fixed amount of material. Stay tuned...

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