## Continuity explained

Imagine you’re driving down a road on your way to pick up a pizza. Ideally, you will be following the exact same path back and forth. If you mark a random point, we'll call it *Point A*, somewhere on your route, you’d expect to cross *Point A* on the way to get your pizza. You'd also expect to cross *Point A *on your way* back* from getting pizza. In addition, if desired you could physically stand on *Point A* and remain stationary (it's not like some sinkhole is at *Point A*!).

If the above is true about *Point A*, we could call our route to get pizza **continuous** at *Point A*. That's because *Point A* is consistent and exists on our route regardless of the direction we go or if we are stationary on it.

Let's now imagine that the route we took to grab pizza was a mathematical function f(x). Using the following criteria, we'd be able to assess whether or not our route to get pizza was continuous at *Point A*!

Let's relate this back to what we said before...

"If the above is true about *Point A*, we could call our route to get pizza **continuous** at *Point A*. That's because *Point A* is consistent and exists on our route regardless of the direction we go or if we are stationary on it."

This whole "regardless of the direction we go" is referring to whether we're going to get the pizza vs. coming back home from getting the pizza.

We can relate going to get the pizza as approaching "x = a" on our function "f(x)" from the left (a^{-}), which is notated as:

We can relate coming back home from getting the pizza as approaching "x = a" on our function "f(x)" from the right (a^{+}), which is notated as:

Now for the second part of what we said before...

"If the above is true about Point A, we could call our route to get pizza **continuous** at Point A. That's because Point A is consistent and exists on our route regardless of the direction we go or if we are stationary on it."

This "stationary" word is referring to if we were actually standing on *Point A*. In other words... does our function "f(x)" have an existing point at "x = a"?

A function f(x) is **continuous** at a given point (x = a) if its limits from the **left-hand** (a^{-}) and **right-hand** (a^{+}) sides are the same value as the **specific** **point** on the function where **x = a**.

## Example #1 - Continuity with regular function

**Example #1**: Verify if the following function is continuous.

To do this, we're going to take the 3 criteria of a continuous point on a function...

...and apply it to this function f(x) at x = 5. (a.k.a. plug in the equation for f(x) into the limit!)

Then, we're going to plug in 5, and see if our values match!

Since all of the numbers agree (at f(5) = 40), then the function is continuous at x = 5!

**Tip**:** **Most basic functions like **linear**, **exponential**, and **polynomial** functions are going to be **continuous** by default. It's typically **piecewise** functions that won't be continuous!

## Example #2 - Continuity with a piecewise function

**Ex. 2: **Verify if the following function is continuous:

Notice this question didn’t specify a point. We have to make that choice on our own here. Looking at the intervals associated to each function, we can guess that the main issue with continuity is going to be x = 0 since that’s the main overlap in the two intervals of -1 < x < 0 and x ≥ 0.

Therefore, we're going to assess the limit here at x = 0!

Like we did before, we're going to take the 3 criteria of a continuous point...

...and apply it to our function f(x) at x = 0!

Now wait... why are we using 2x + 5 for "1." and x^{2} in "2." and "3."? Shouldn't we be using the same function?

### When to use f(x) = 2x + 5

In #1, we're approaching 0 from the left-hand side (a.k.a. 0^{-}).

This means that we'd be approaching 0 from numbers *less than 0*. And based on our prompt, when x is less than 0 (a.k.a. "-1 < x < 0")...

...f(x) is represented by the 2x + 5 function!

### When to use f(x) = x^{2}

In #2, we're approaching 0 from the right-hand side (a.k.a. 0^{+}).

This means that we'd be approaching 0 from numbers *greater than 0*. And based on our prompt, when x is *greater than* or equal to 0 (a.k.a. "x ≥ 0")...

...f(x) is represented by the x^{2} function!

In #3, we're simply assessing f(x) at the point x = 0. Based on the prompt, when x is greater than or *equal to* 0 (a.k.a. "x ≥ 0")...

...f(x) is represented by the x^{2} function!

Therefore, at the exact point of x = 0, we'll be using x^{2} function!

### Plugging in x = 0

Now that we've figured out why our piecewise functions occur where they do, let's go ahead and plug in 0 for x with direct substitution.

Since our numbers do not all equal each other (we're getting 5 and 0)...

...we can conclude that this function is not continuous at x = 0!

A function f(x) is **not continuous** at a given point (x = a) if its limits from the **left-hand** (a^{-}) and **right-hand** (a^{+}) sides or the value at the **specific** **point** on the function where x = a are **different**.

## Exam 1 Cram Kit

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Apply | PRACTICE EXAM 1 (PREVIEW ONLY) | |

Concept | Limits | |

Concept | Continuity | |

Concept | What is a derivative? (PREVIEW ONLY) | |

Concept | Evaluating Derivatives (PREVIEW ONLY) |