### Motivating Questions

- What does it mean to say that a function \(f\) is continuous at \(x = a\text{?}\)
- What does it mean to say that a function \(f\) is continuous everywhere?

- What does it mean to say that a function \(f\) is continuous at \(x = a\text{?}\)
- What does it mean to say that a function \(f\) is continuous everywhere?

The main topics of this section are also presented in the following videos:

- Introduction to Continuity
^{ 1 }`unl.yuja.com/V/Video?v=7114243&node=34303294&a=154632527&autoplay=1`

- Continuity
^{ 2 }`unl.yuja.com/V/Video?v=7114241&node=34303318&a=58125516&autoplay=1`

- Piecewise Functions
^{ 3 }`unl.yuja.com/V/Video?v=7114242&node=34303228&a=214181532&autoplay=1`

- Intermediate Value Theorem
^{ 4 }`unl.yuja.com/V/Video?v=7114240&node=34303312&a=201475606&autoplay=1`

- Continuity at a Point
^{ 5 }`unl.yuja.com/V/Video?v=7114239&node=34303220&a=120711774&autoplay=1`

In this section we introduce the idea of a continuous function. Many of the results in calculus require that the functions be continuous, so having a strong understanding of continuous functions will be very important.

Intuitively, a function is continuous if we can draw its graph without ever lifting our pencil from the page. Alternatively, we might say that the graph of a continuous function has no jumps or holes in it.

First consider the function in the left-most graph above. Note that \(f(1)\) is not defined, which leads to the resulting hole in the graph of \(f\) at \(a = 1\text{.}\) If you were to draw the graph of \(f\) yourself then you would need to lift your pencil when you reached \(f(1)\text{.}\) We will naturally say that \(f\) is *not continuous at \(a = 1\)*. For the function \(g\text{,}\) we observe that while \(g(1)\) is defined, the value of \(g(1) = 2\) is not what you would “expect.” Specifically, you would expect \(g(1)\) to be 3, not 2. Thus, to draw the graph of \(g\) you would need to lift your pencil at \(a=1\text{.}\) Again, we will say that \(g\) is *not continuous* at \(a=1\text{,}\) even though the function is defined at \(a =
1\text{.}\) Finally, the function \(h\) appears to be the most “well-behaved” of all three, since at \(a = 1\) the function value is what you might expect it to be if you were to try and draw the graph of the function without lifting your pencil. In this case we would say that \(h\) is continuous at \(a=1\text{.}\)

The above examples demonstrate a discontinuity commonly know as a *removable discontinuity*. This is, however, not the only way in which a function can be discontinuous. Another type of discontinuity is the so-called *jump* discontinuity illustrated below.

A third type of discontinuity is the so-called *infinite* discontinuity. Infinite discontinuities exist at points where the values of a function diverge to infinity. A classic example of an infinite discontinuity is the point \(x=0\) for the function \(f(x)=\frac{1}{x}\text{;}\) you can see the behavior of the infinite discontinuity in the graph of \(y=f(x)\) below.

Consider the function

\begin{equation*}
f(x)=\frac{3x(x+4)}{x^2-16}\text{.}
\end{equation*}

Are there \(x\)-values where the function is discontinuous? If so, what are those \(x\)-values and why type of discontinuity occurs at that value?

Consider the graph of \(y=f(x)\text{.}\)

\(f(x)\) is discontinuous at \(x=4, -4\text{;}\) however, the discontinuity at \(x=-4\) is removable.

We first consider a graph of \(y=f(x)\text{,}\) shown below.

Visual inspection of the graph certainly indicates discontinuities. However, we can make our visual inspection more precise through a little algebra. We start by expanding the numerator and denominator of the function. Specifically, we have

\begin{equation*}
f(x)=\frac{3x(x+4)}{x^2-16}=\frac{3x(x+4)}{(x+4)(x-4)}.
\end{equation*}

From here we see that there are two \(x\)-values at which the function is undefined: at \(x=-4,4\text{.}\) Since the term \(x+4\) occurs in both the numerator and denominator, the point \(x=-4\) is a removable discontinuity. Since the term \(x-4\) occurs only in the denominator, \(x=4\) is an infinite discontinuity.

In many cases a simple function like \(f(x)=x^2\) may not fully describe the behavior of a phenomenon. In some of these cases we can turn to *piecewise* functions to give us the tools we need.

A piecewise-defined function is one which is defined not by a single equation, but by two or more. Each equation is valid for some interval.

Graph the function defined by

\begin{equation*}
f(x) = \begin{cases} x +1 \amp \text{if } x\le 1\\ 3 \amp \text{if } x\gt 1
\end{cases}.
\end{equation*}

Try graphing each part of the piecewise funtion separately on the indicated domains.

Think of the plane as divided into two regions by the vertical line \(x = 1\text{,}\) as shown in Figure 1.8. In the left-hand region (\(x \le 1\)), we graph the line \(y = x + 1\text{.}\) (The fastest way to graph the line is to plot its intercepts, \((-1, 0)\) and \((0, 1)\text{.}\))

Notice that the value \(x = 1\) is included in the first region, so \(f
(1) = 1 + 1 = 2\text{,}\) and the point \((1, 2)\) is included on the graph. We indicate this with a solid dot at the point \((1, 2)\text{.}\) n the right-hand region (\(x \gt 1\)), we graph the horizontal line \(y
= 3\text{.}\) The value \(x = 1\) is not included in the second region, so the point \((1, 3)\) is not part of the graph. We indicate this with an open circle at the point \((1, 3)\text{.}\)

The jumps that a piecewise function possesses make piecewise functions a natural place in which to explore continuity.

The function

\begin{equation*}
h(x)=\begin{cases} 4x^2 \amp \text{ if } x\leq 1 \\ 3x+3 \amp \text{ if } x\gt1
\end{cases}
\end{equation*}

is discontinuous at \(x=1\text{.}\) It is continuous on any interval that does not include \(x=1\text{.}\)

Not only can we ask questions about when a piecewise function is continuous, we can also ask questions about how to make a piecewise function continuous by varying parameters.

Consider the piecewise function

\begin{equation*}
D(x)=\begin{cases} 4x^2-k\amp\text{ if } x \lt 2,\\ kx+1\amp\text{ if } x
\geq 2.\\ \end{cases}
\end{equation*}

Find the value of \(k\) to make this function continuous for all \(x\text{.}\)

The point where the function may not be continuous is at the potential “jump”-point. Specifically, when \(x=2\text{.}\)

\(k=5\)

The point where the function may not be continuous is at the potential “jump”-point. Specifically, when \(x=2\text{.}\) For the function to be continuous we need the two “branches” of the function to meet when \(x=2\text{.}\) That is we need \(4\cdot(2)^2-k=k\cdot(2)+1\) which is equivalent to \(k=5\text{.}\) The figure below shows a graph of \(y=D(x)\) for \(k=5\text{.}\)

Up to this point we have talked about the continuity of a function at a point. Before stating the Intermediate Value Theorem we begin by defining a continuous function. We will formalize this definition in Section 1.2 with the introduction of limits.

A function is said to be continuous on an interval \((a,b)\) if the function has no points of discontinuity on that interval.

One of the most important theorems in Calculus is the Intermediate Value Theorem, which we state formally below.

Suppose \(f\) is continuous on a closed interval \([a,b]\text{.}\) If \(k\) is any number between \(f(a)\) and \(f(b)\text{,}\) then there is at least one number \(c\) in \([a,b]\) such that \(f(c)=k\text{.}\)

The figure below illustrates the theorem. The function \(f(x) \text{,}\) whose graph is drawn in black, is continuous on the closed interval \([a,b] \text{.}\) Suppose we choose a number \(k \) between \(f(a) \) and \(f(b) \text{.}\) The Intermediate Value Theorem says that the horizontal line \(y = k \) must intersect the graph of \(f(x) \) in at least one point.

In the figure, the line \(y = k \) intersects the graph of \(f(x) \) three times. The \(x \)-coordinate of any one of these points of intersection is a number \(c \) such that \(f(c) = k \text{.}\)

Amber measures herself before going to bed and is 67 inches tall. She measures herself again when she wakes up 8 hours later, and is now 68 inches tall. Let \(h(t)\) be Amber’s height in inches \(t\) hours after the first measurement. Then \(h(0)=67\) and \(h(8)=68\text{.}\)

Do you think there is a time between when Amber goes to bed and when she wakes up that she is 67.5 inches tall?

Since Amber is likely growing continuously (i.e., she did not jump from 67 inches to 68 inches instantly), the Intermediate Value Theorem holds for the height function \(h(t)\text{.}\) The Intermediate Value Theorem states that at some time while Amber slept, she measured 67.5 inches tall.

Consider the function \(f(x)=x^3-3x^2+3x-4\text{.}\) Use the Intermediate Value Theorem to show that \(f(x)\) has a root between \(x=2\) and\(x=3\text{.}\)

What are the values of \(f(2)\) and \(f(3)\text{?}\)

We want to show that there is some number \(c\) between 2 and 3 such that \(f(c)=0\text{.}\) We consider the value of the function \(f\) at \(x=2\) and \(x=3\text{:}\)

\begin{equation*}
f(2)=8-12+6-4=-2\lt 0
\end{equation*}

and

\begin{equation*}
f(3)=27-27+9-4=5\gt 0.
\end{equation*}

Thus \(f(2)\lt 0\lt f(3)\text{;}\) in other words, 0 is a number between \(f(2)\) and \(f(3)\text{.}\) The function \(f\) is a polynomial, so it is a continuous function. Thus the Intermediate Value Theorem says there is a number \(c\) between 2 and 3 such that \(f(c)=0\text{.}\) In other words, \(f\) has a root \(c\) in the interval \((2,3)\text{.}\)

- Introduction to Continuity
^{ 7 }`app.vidgrid.com/view/sWXRwqF0NkCm`

- Continuity
^{ 8 }`app.vidgrid.com/view/Pho0yVX1l2qf`

- Piecewise Functions
^{ 9 }`app.vidgrid.com/view/t7gzHYWGqlRp`

- Intermediate Value Theorem
^{ 10 }`app.vidgrid.com/view/rMY8eKffe0P9`

- Continuity at a Point
^{ 11 }`app.vidgrid.com/view/Y34X1mPC19WN`

Determine the point at which the function \(\displaystyle f(x)= \frac {1} {x-2}\) is discontinuous and state the type of discontinuity: removable, jump, infinite, or none of these.

\(x=\)

- Choose the type

Determine the points at which the function is discontinuous and state the type of discontinuity: removable, jump, infinite, or none of these.

\(\displaystyle f(x)= \frac {x - 2} {\mid x - 1 \mid}\)

\(x=\)

- Choose the type

Enter T or F depending on whether the statement is true or false. (You must enter T or F -- True and False will not work.)

- The function \(\displaystyle f(t)= \frac {1} {t^2-t}\) has an infinite discontinuity at \(t=1\text{.}\)

Answer the questions based on the graph of \(y = f(x)\) shown above.

If you are having a hard time seeing the picture clearly, click on the picture. It will expand to a larger picture on its own page so that you can inspect it more clearly.

a) \(f\)

- is
- is not

continuous at \(-2\text{.}\)

b) \(f\)

- is
- is not

continuous at \(-1\text{.}\)

c) \(f\)

- is
- is not

continuous at \(0\text{.}\)

d) \(f\)

- is
- is not

continuous at \(1\text{.}\)

e) \(f\)

- is
- is not

continuous at \(3\text{.}\)

f) \(f\)

- is
- is not

continuous at \(4\text{.}\)

Answer the questions based on the graph of \(y = f(x)\) shown above.

If you are having a hard time seeing the picture clearly, click on the picture. It will expand to a larger picture on its own page so that you can inspect it more clearly.

a) \(f\)

- is
- is not

continuous at \(0\text{.}\)

b) \(f\)

- is
- is not

continuous at \(1\text{.}\)

c) \(f\)

- is
- is not

continuous at \(2\text{.}\)

An electrical circuit switches instantaneously from a 6 volt battery to a 21 volt battery 7 seconds after being turned on. Sketch on a sheet of paper a graph the battery voltage against time. Then fill in the formulas below for the function represented by your graph.

For \(t \lt \) , \(V(t) =\)

For \(t \ge\) , \(V(t) =\) .

At what point or points is your function discontinuous?

\(t =\)

Find \(k\) so that the following function is continuous:

\begin{equation*}
f(x) = \begin{cases}
kx \amp \hbox{if}\quad 0\le x \lt 4 \\
5 x^2 \amp \hbox{if}\quad 4\le x.
\end{cases}
\end{equation*}

\(k =\)

Find a value of the constant \(k\text{,}\) if possible, at which

\begin{equation*}
f(x)=\begin{cases}kx^{2} \amp x\leq5 \\ -9x+k \amp x>5 \end{cases}
\end{equation*}

is continuous everywhere.

\(k=\) (enter "none" if no value).

If possible, choose \(k\) so that the following function is continuous on any interval:

\begin{equation*}
f(x)= \begin{cases} \frac{8x^{5}-16x^{4}}{x-2}\quad \amp x \ne 2 \\
k \amp x = 2. \end{cases}
\end{equation*}

\(k =\)

Consider the function \(f(x) = 2 x^3 + 3 x^2 + 7\text{.}\) For what values of \(k\) does the Intermediate Value Theorem tell us that there is a \(c\) in the interval \([0,1]\) such that \(f(c) = k\text{?}\)

\(\le k \le\) .