Question Video: Discussing the Continuity of a Constant Function | Nagwa Question Video: Discussing the Continuity of a Constant Function | Nagwa

# Question Video: Discussing the Continuity of a Constant Function Mathematics • Second Year of Secondary School

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Discuss the continuity of the function π(π₯ ) = 4.

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### Video Transcript

Discuss the continuity of the function π of π₯ is equal to four.

In this question, weβre given a function π of π₯ and weβre asked to discuss the continuity of this function. To do this, we first need to know exactly what the question means by discussing the continuity of a function. Whenever weβre asked to discuss the continuity of a function, what this is really telling us to do is find all of the values of π₯ where our function π of π₯ is continuous. And usually, weβll give our answer as a set, the set of values of π₯ where π of π₯ is continuous. And in fact thereβs a few different ways of answering this question. The easiest way to answer this question is to recall a couple of facts about continuous functions.

We just need to recall two things. First, we know all polynomials are continuous across the entire set of real numbers. Next, we also know that the function π of π₯ is equal to four is a polynomial. However, we could also use something simpler. All constant functions are continuous on the set of real numbers, and the function π of π₯ is equal to four is a constant function. And both of these give us the same answer to our question. π of π₯ is continuous on the set of real numbers because it is a constant function or, alternatively, because itβs a polynomial. And although this answers the question, it doesnβt help us exactly know why our function π of π₯ is continuous. So instead, letβs do this directly from the definition of continuity.

We recall that we say a function π of π₯ is continuous at a value of π₯ is equal to π if the following three conditions are true. First, π evaluated at π must exist. Another way of saying this is π is in the domain of our function π. Next, we need the limit as π₯ approaches π of π of π₯ to exist. Finally, we need the limit as π₯ approaches π of π of π₯ to be equal to π evaluated at π. And sometimes youβll see this definition written only with the third step because the first two steps are sometimes implied by the third step. For example, if π evaluated at π did not exist, then itβs hard to say the limit as π₯ approaches π of π of π₯ is equal to something which doesnβt exist. However, weβll write this out in full and check all three of these conditions.

We want to show that our function π of π₯ is continuous for all real values of π. So letβs start by letting π be any real number. We need to show all three of these conditions are true for π of π₯. First, weβre going to need to find the value of π evaluated at π. To find π evaluated at π, we need to substitute π₯ is equal to π into our definition of the function π of π₯. Of course, π of π₯ is the constant function four, so this is always just going to be equal to four. So our first condition is true. π of π is always equal to four. Another way of saying this is the domain of π of π₯ is all real numbers.

Letβs now move on to our second condition. We need to show the limit as π₯ approaches π of π of π₯ exists, and thereβs a lot of different ways we could do this. All of them will give us the same answer. The easiest way to do this is to consider the definition of a limit. We recall, we say the limit as π₯ approaches π of π of π₯ is equal to some finite value of πΏ if the values of π of π₯ approach πΏ as the values of π₯ approach π from both sides. And in our case, we can actually evaluate this limit for any value of π.

First, recall that our function π of π₯ is just equal to the constant four. And if our function π of π₯ is the constant four, then the outputs of our function π of π₯ are always equal to four. So the outputs of our function are constant; they donβt change regardless of our input value of π₯. So our values of π₯ donβt even matter, will always output in the value of four. So our limit is always going to approach the value of four regardless of our value of π. Therefore, weβve shown for any real value π, the limit as π₯ approaches π of π of π₯ does exist. In fact, itβs equal to four. And itβs also worth pointing out we couldβve done it just by using the fact that the limit of a constant is equal to the constant itself. Either method would work and they give us the same answer.

The last thing we need to check is our third and final definition of continuity. However, weβve actually already shown that this is true. When we were proving the first part of continuity, we showed that π evaluated at π is equal to four, and when we were proving the second part of continuity, we showed that the limit as π₯ approaches π of π of π₯ is also equal to four. So both of these are equal to four. So all three of our parts of continuity are true for π of π₯ for any value of π. And therefore, weβve shown directly from the definition of continuity that our function π of π₯ is continuous for all real values of π₯.

In this question, we were able to show two different ways of proving the continuity of our function π of π₯. We were able to show that π of π₯ is continuous on the set of real numbers because it is a constant function.

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