1.04 Continuity, end behavior, and the intermediate value theorem

Continuity

Recall that the domain of a function is the set of $x$x values on which the function is defined. Graphically, we can think of the domain as all values of $x$x which correspond to a point on the curve.

A function $f\left(x\right)$f(x) is said to be continuous at a point $x=c$x=c if:

  i. the function is defined at that point (that is, $c$c is in the domain of $f\left(x\right)$f(x)), and

  ii. we can get function values as close as we like to $f\left(c\right)$f(c) by taking a small enough region of domain values around $x=c$x=c.

A function $f\left(x\right)$f(x) is then said to be continuous over an interval if it is continuous at all points in that interval. A continuous function is a function if its domain is a single interval and it is defined at every point in that interval.

If instead there exists a point in the function's domain at which it is not continuous, we say that the function is discontinuous at that point.

Let's now take a look at some different types of continuous functions by inspecting their graphs.

 

Continuous functions over the real numbers

Linear functions (that is, functions of the form $f\left(x\right)=ax+b$f(x)=ax+b) are defined for all real numbers. We can represent this by saying its domain is the interval $\left(-\infty,\infty\right)$(−∞,∞).

Look at the linear function in the graph above. At the domain value $x=2$x=2 we have the function value $f\left(2\right)=3$f(2)=3. We can get other function values as close as we like to $3$3 by taking a small enough region of the domain around $x=2$x=2. Watch the applet below for a demonstration.

Created with Geogebra

There is nothing special about $x=2$x=2 in this case - the same is true for any other point in the domain! So the function above is a continuous function, as are all linear functions.

 

Other examples of continuous functions which have a domain of $\left(-\infty,\infty\right)$(−∞,∞) include any type of polynomial function (quadratic functions, cubic functions etc) as well as exponential functions.

Some of these types are sketched below:

 

Continuous functions over smaller domains

Many functions are not defined for every real number, but are still continuous at every point in their domain, which is a single interval. Let's take a look at some of these types of functions.

A logarithmic function, which takes a form such as $f\left(x\right)=a\log\left(x-h\right)+k$f(x)=alog(x−h)+k, is only defined when the input of the logarithm is positive.

The logarithmic function shown in the graph above has a vertical asymptote at $x=1$x=1. It is defined on all $x$x values larger than $1$1, and we can represent this domain as $\left(1,\infty\right)$(1,∞). Its domain is a single interval, and it is continuous at all points in this domain, so it is also a continuous function.

 

A square root function, which takes a form such as $f\left(x\right)=a\sqrt{x-h}+k$f(x)=a√x−h+k, is defined when the input of the square root is not negative.

The function $f\left(x\right)=\sqrt{x+4}-1$f(x)=√x+4−1 has been graphed above. The value under the square root cannot be negative, and (as we can see from the graph) this results in a domain of $\left[-4,\infty\right)$[−4,∞). Its domain is a single interval and the function is continuous at all points in this domain (including the endpoint where $x=-4$x=−4), and so it is also a continuous function.

 

Discontinuous functions

As we have seen so far, most of the functions that we are familiar with are continuous. Not all functions are continuous though! Functions can have many different types of discontinuity.

Let's take a look at an example of a step function:

In the graph above, the step function has a step at $x=0$x=0. It takes a constant value (in this case $-1$−1) for $x<0$x<0 and a different constant value (in this case $+1$+1) for $x\ge0$x≥0, and so its domain is $\left(-\infty,\infty\right)$(−∞,∞) (that is, all of the real numbers).

This function has a point of discontinuity at $x=0$x=0 - no matter how small we make our region of domain values around $x=0$x=0, we will always include part of the function that takes the value $-1$−1 and part of the function that takes the value $+1$+1. This means that the step function is not a continuous function. It has what is known as a jump discontinuity.

 

A rational function has a form such as $f\left(x\right)=\frac{a}{x-h}+k$f(x)=ax−h​+k. Since the domain variable $x$x appears in the denominator there is a value of $x$x for which the function is not defined, so this value of $x$x is not in the domain of the function.

The rational function in the graph above has a vertical asymptote at $x=1$x=1. The domain of this function is all of the real numbers except for $x=1$x=1, which can be written as $\left(-\infty,1\right)\cup\left(1,\infty\right)$(−∞,1)∪(1,∞). The domain of the rational function is not a single interval. Therefore, it is not a continuous function. Its discontinuity is classified as an infinite discontinuity.

A function has removable discontinuity if the function is continuous everywhere except for a hole at a point where $x=c$x=c. Notice that in the next applet, domain values close to $f\left(x\right)=2$f(x)=2 do not correspond to function values close to $f\left(2\right)=5$f(2)=5. So we say that this function is discontinuous.

Created with Geogebra

 

Practice questions

Question 1

Question 2

 

Asymptotes

An asymptote of a function is a straight line which the function values approach under certain conditions. There are three types of asymptotes: horizontal, vertical and oblique.

Horizontal asymptotes have equations of the form $y=c$y=c. They occur when the function approaches a constant value $c$c as $x$x tends towards positive or negative infinity.

Vertical asymptotes have equations of the form $x=c$x=c. They occur when the function values tend towards positive or negative infinity as $x$x approaches the constant value $c$c.

It is also possible for a function to have an oblique asymptote or slanted asymptote. Graphically, this means that the function approaches a straight line with a non-zero slope as $x$x tends towards positive or negative infinity.

As an example, here is the graph of a rational function:

Notice that the function values tend towards $\pm\infty$±∞ as the values of $x$x approach $1$1. This function has a vertical asymptote at $x=1$x=1, which has been displayed as a dashed line.

There is also a horizontal dashed line with equation $y=2$y=2, which is the horizontal asymptote of this function. We can see that the function values approach $2$2 as the values of $x$x tend towards positive and negative infinity.

 

 

Practice questions

Question 3

Question 4

 

Intermediate value theorem

If a function is continuous, we can use the intermediate value theorem to locate the zeros of the function.

 

Finding zeros

One immediate implication of the intermediate value theorem is that it gives us a way to determine whether a continuous function has a zero in a given interval. Notice that if the sign of $f(a)$f(a) is different to the sign of $f(b)$f(b), then there must be some function value between $f(a)$f(a) and $f(b)$f(b) that is exactly equal to $0$0. There will then be a corresponding $x$x-value for this function value that we call the zero of the function.

Exploration

A certain quantity varies with time according to the rule $x(t)=t^3-2t^2+1$x(t)=t3−2t2+1. It is clear that $x(1)=0$x(1)=0, but we wonder whether there are other values of $t$t such that $x(t)=0$x(t)=0.

We choose some values for $t$t. Say, $t=-1$t=−1, $t=0.5$t=0.5, $t=1.5$t=1.5, and $t=2$t=2, and we calculate $x(t)$x(t) in each case.

$t$t	$-1$−1	$0.5$0.5	$1.5$1.5	$2$2
$x\left(t\right)$x(t)	$-2$−2	$0.625$0.625	$-0.125$−0.125	$1$1

By the intermediate value theorem, we see that the function $x(t)$x(t) has a zero between $t=-1$t=−1 and $t=0.5$t=0.5, another between $0.5$0.5 and $1.5$1.5 (which we already knew about), and another between $t=1.5$t=1.5 and $t=2$t=2.

We conclude that there is at least three zeros in the interval between $t=-1$t=−1 and $t=2$t=2.

Practice questions

Question 5

Question 6

Consider the polynomial $P\left(x\right)=4x^3-x^2+7x+7$P(x)=4x3−x2+7x+7. Sharon would like to know if it has a real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7.

1. Find $P\left(-0.8\right)$P(−0.8) to one decimal place.

2. Find $P\left(-0.7\right)$P(−0.7) to one decimal place.

3. What conclusion can Sharon make about $P\left(x\right)$P(x) between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7 using the intermediate value theorem?

   There is exactly one real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7

   A

   She cannot conclude anything about the zeros of the function.

   B

   There is at least one real zero between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7

   C

   There are no real zeros between $x=-0.8$x=−0.8 and $x=-0.7$x=−0.7

   D

Reveal SolutionWatch video

Question 7