What three quantities must agree at the boundary between the pieces of a piecewise function for it to be continuous there?

The value of the expression defining the function to the left of the boundary, the value of the expression to the right of it, and the value of the function at the boundary itself. That is the same three-part definition applied to a function given in pieces, not a separate rule.

Define a jump discontinuity .

A jump discontinuity occurs at a point where the limit from the left and the limit from the right both exist but are not equal. The graph makes a sudden leap between two values there.

Define an essential discontinuity .

An essential discontinuity , also called an infinite discontinuity, occurs at a point where the limit from the left or the limit from the right, or both, either fails to exist or is infinite. A vertical asymptote is the usual way this happens.

How do you tell a jump discontinuity from an essential one?

Look at whether the two one-sided limits exist as finite numbers . If both do and they simply disagree it is a jump, whereas if either one fails to exist or runs off to infinity it is essential.

True or False? The intermediate value theorem can be used on a function for which you have no formula.

True. The only things needed are continuity on a closed interval and the two end values, and both of those can come from a table or from the wording of a question. It is an existence result, so it never asks you to work the function out anywhere else.

Why does the intermediate value theorem fail for a function with a jump discontinuity?

Because the graph can leap straight over a value, taking one below it on the left and one above it on the right without ever taking the value itself. Continuity on the closed interval is what rules that out, which is why it is a hypothesis of the theorem rather than a detail.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 1: Limits & ContinuityContinuity

Continuity (College Board AP® Calculus BC): Flashcards

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FrontContinuity
Fill in the three missing parts of the definition of continuity at a point:
A function $f$ is continuous at $x = c$ provided that $______$ exists, that $______$ exists, and that these two are $______$ to each other.

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Cards in this collection (26)

Fill in the three missing parts of the definition of continuity at a point:
A function $f$ is continuous at $x = c$ provided that $______$ exists, that $______$ exists, and that these two are $______$ to each other.
The completed definition is: $f$ is continuous at $x = c$ provided that $f (c)$ exists, that $\lim_{x \to c} f (x)$ exists, and that these two are equal to each other.
All three parts have to hold, and each of them can fail on its own.
For $g (x) = \frac{x}{x}$ the limit at $0$ is $1$ . Which part of the definition of continuity fails at $x = 0$ ?
The first part: $g (0) = \frac{0}{0}$ is not defined, so there is no value there for the limit to be compared with.
A limit existing is not enough on its own, since the function has to take a value at the point as well.
Define what it means for a function to be continuous over an interval.
A function is continuous over an interval when it is continuous at every point in that interval.
So $g (x) = \frac{x}{x}$ is not continuous over $[- 1, 1]$ , but it is continuous over any interval that leaves $0$ out.
A function $h$ has $h (0) = 1$ and $\lim_{x \to 0} h (x) = 0$ . Is $h$ continuous at $x = 0$ ?
No, it is not.
Both the value and the limit exist, but they are not equal, and the definition needs all three of its parts to hold.
A function can be perfectly well defined at a point and still be discontinuous there.
True or False?
A function that is not continuous over all the real numbers can still be a continuous function.
True.
A function is called continuous when it is continuous at every point of its own domain, and points outside the domain are never asked about.
Defining $j (x) = \frac{1}{x}$ for $x \neq 0$ makes $j$ a continuous function, even though its graph comes in two separate branches.
What three quantities must agree at the boundary between the pieces of a piecewise function for it to be continuous there?
The value of the expression defining the function to the left of the boundary, the value of the expression to the right of it, and the value of the function at the boundary itself.
That is the same three-part definition applied to a function given in pieces, not a separate rule.
If $f$ and $g$ are continuous at $x = a$ , which combinations of them are too?
All four of $f + g$ , $f - g$ , $f \cdot g$ and $\frac{f}{g}$ are continuous at $a$ , with the quotient needing $g (a) \neq 0$ .
That condition is there because dividing by zero leaves the quotient with no value at $a$ , so its continuity cannot even be tested.
The function $f$ is $x + b$ for $x \leq 3$ and $b x^{2}$ for $x > 3$ , and it is continuous. Find $b$ .
The value is $b = \frac{3}{8}$ .
Continuity at $x = 3$ forces the two expressions to agree there, so $3 + b = 9 b$ , which gives $3 = 8 b$ .
Define a removable discontinuity.
A removable discontinuity is a point $x = c$ where $\lim_{x \to c} f (x)$ exists, but $f (c)$ either does not exist or is not equal to that limit.
On a graph it shows as a hole, with the rest of the curve arriving at the missing point from both sides.
A removable discontinuity can arise in two ways. What are they?
Either $f (c)$ does not exist at all, leaving an empty hole in the graph, or $f (c)$ does exist but sits away from the hole at a value different from the limit.
The second kind is easy to miss, because the graph really does pass through a point at that value of $x$ .
Fill in the missing expression that removes a discontinuity at $x = c$ :
$f (c) =______$
The completed definition is $f (c) = \lim_{x \to c} f (x)$ .
Once the value at the point matches the limit, every part of the definition of continuity holds there and the function is continuous.
True or False?
Every discontinuity at which the function is undefined can be removed.
False.
Being undefined is not the test; what matters is whether the limit exists at that point.
The function $\frac{1}{x}$ is undefined at $0$ and has no limit there, so no choice of value for $f (0)$ would make it continuous.
The function $f$ is defined by $f (x) = \frac{x^{2} + 3 x}{x}$ . How can its discontinuity at $x = 0$ be removed?
Factor and simplify to get $\frac{x (x + 3)}{x} = x + 3$ , so $\lim_{x \to 0} f (x) = 3$ .
Define $f (0) = 3$ , keeping $\frac{x^{2} + 3 x}{x}$ for every other value of $x$ , and the function becomes continuous.
When a removable discontinuity is removed by redefining $f (c)$ , what has actually changed about the function?
Only the single value $f (c)$ , which is now set equal to the limit there.
Every other value is untouched and the limit at $c$ was already correct, so nothing but that one point has moved.
Define a jump discontinuity.
A jump discontinuity occurs at a point where the limit from the left and the limit from the right both exist but are not equal.
The graph makes a sudden leap between two values there.
A function is $x^{3} - x^{2} - 2 x$ for $x < 1$ and $8 - x^{2} - 3 x$ for $x \geq 1$ . What kind of discontinuity does it have at $x = 1$ ?
A jump discontinuity.
The limit from the left is $1 - 1 - 2 = - 2$ and the limit from the right is $8 - 1 - 3 = 4$ , so both exist and are finite but they disagree.
True or False?
Giving the function a new value at the point can turn a jump discontinuity into a continuous point.
False.
Changing $f$ at the single point cannot shift either one-sided limit, and it is those two that disagree.
A discontinuity can only be repaired where the two-sided limit exists, which at a jump it does not.
Define an essential discontinuity.
An essential discontinuity, also called an infinite discontinuity, occurs at a point where the limit from the left or the limit from the right, or both, either fails to exist or is infinite.
A vertical asymptote is the usual way this happens.
A function is $x - 3$ for $x \leq 5$ and $4 - \frac{1}{x - 5}$ for $x > 5$ . What kind of discontinuity does it have at $x = 5$ ?
An essential discontinuity.
The limit from the left is $2$ , which is also the value $f (5)$ , but from the right $\frac{1}{x - 5}$ grows without bound, so that one-sided limit is $- \infty$ .
How do you tell a jump discontinuity from an essential one?
Look at whether the two one-sided limits exist as finite numbers.
If both do and they simply disagree it is a jump, whereas if either one fails to exist or runs off to infinity it is essential.
State the intermediate value theorem.
If $f$ is continuous on the closed interval $[a, b]$ and $d$ is a number between $f (a)$ and $f (b)$ , then there is at least one number $c$ between $a$ and $b$ with $f (c) = d$ .
In plainer terms, a continuous function takes every value between its two end values somewhere along the interval.
True or False?
The intermediate value theorem can be used on a function for which you have no formula.
True.
The only things needed are continuity on a closed interval and the two end values, and both of those can come from a table or from the wording of a question.
It is an existence result, so it never asks you to work the function out anywhere else.
Fill in the missing conclusion about the zeros of a continuous function:
If a continuous function is positive at one $x$ -value and negative at another, then it has at least one $______$ between them.
The completed statement is: if a continuous function is positive at one $x$ -value and negative at another, then it has at least one zero between them.
This is the intermediate value theorem with $d = 0$ , since $0$ lies between any positive value and any negative one.
A continuous function has $f (1) = 5$ and $f (3) = 7$ . What does the intermediate value theorem say about solutions of $f (x) = 6$ and of $f (x) = 9$ on $[1, 3]$ ?
For $f (x) = 6$ it guarantees at least one solution with $1 < x < 3$ , because $6$ lies between $5$ and $7$ .
For $f (x) = 9$ it says nothing at all, because $9$ lies outside that range, and silence is not the same as ruling a solution out.
Why does the intermediate value theorem fail for a function with a jump discontinuity?
Because the graph can leap straight over a value, taking one below it on the left and one above it on the right without ever taking the value itself.
Continuity on the closed interval is what rules that out, which is why it is a hypothesis of the theorem rather than a detail.
What does the intermediate value theorem not tell you about the value $c$ ?
It does not tell you where $c$ is, only that at least one such number exists somewhere between $a$ and $b$ .
Nor does it tell you how many such numbers there are, since there may well be several.

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