# Question Video: Finding the Values of the Variables Which Make a Piecewise-Defined Function Continuous at a Certain Point Mathematics • Higher Education

Let 𝑓(𝑥) = 𝑎𝑥² + 18 if 𝑥 ≠ 9, 𝑓(𝑥) = −6𝑎 if 𝑥 = 9. Find the value of 𝑎 that makes 𝑓 continuous at 𝑥 = 9.

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### Video Transcript

Let 𝑓 of 𝑥 equal 𝑎𝑥 squared plus 18 if 𝑥 is not equal to nine and minus six 𝑎 if 𝑥 equals nine. Find the value of 𝑎 that makes 𝑓 continuous at 𝑥 equals nine.

This question is asking us about the continuity of a function at a point, so let’s write down the definition of continuity as a point.

A function is continuous as a point 𝑥 equals 𝑐 if firstly 𝑓 of 𝑐 exists. Secondly, the limit is 𝑥 tends to 𝑐 of 𝑓 of 𝑥 exists. And thirdly, these two things are equal. So the limit as 𝑥 tends to 𝑐 of 𝑓 of 𝑥 is equal to the value of the function there, 𝑓 of 𝑐. In our question, the value of 𝑐 is nine. And so we have to find our value of 𝑎 for which 𝑓 of nine exists, first of all. Secondly, that the limit as 𝑥 tends to nine of 𝑓 of 𝑥 exists. And thirdly that these two things are equal.

So let’s go through these steps one by one. Our first task is to show that 𝑓 of nine exists. Looking now at the definition of our function, we see that 𝑓 of 𝑥 is equal to negative six 𝑎 if 𝑥 is equal to nine. Another way of writing that is to say that 𝑓 of nine is equal to negative six 𝑎. And this of course exists for any value of 𝑎.

Moving on to our second step, we have to find the values of 𝑎 for which the limit as 𝑥 tends to nine of 𝑓 of 𝑥 exists. And we do this by just finding this limit. The value of this limit only depends on values of the function for which 𝑥 is not equal to nine; 𝑥 is tending to nine but never actually equals nine in the limit. And so we can use the definition of the function when 𝑥 is not equal to nine, which is 𝑎𝑥 squared plus 18. And so we get this line of working here, a limit as 𝑥 tends to nine of 𝑓 of 𝑥 is equal as the limit as 𝑥 tends to nine of 𝑎𝑥 squared plus 18.

Now whatever the value of 𝑎 is, 𝑎𝑥 squared plus 18 is just some polynomial. And we know that as polynomials are continuous, the limit of the polynomial is just equal to the value of the polynomial about limiting point. So we can substitute the value nine for 𝑥 and we get 𝑎 times nine squared plus 18. And we can simplify that to 81 𝑎 plus 18. That’s our limit in terms of 𝑎.

And we can see that we’ve completed our second step pretty much, because whatever the value of 𝑎 is of course this limit exists. So now we found both 𝑓 of nine and the limit as 𝑥 tends to nine of 𝑓 of 𝑥, both of those in terms of 𝑎. Now we have to find the value of 𝑎 for which these two things are equal. This is just a matter of solving this linear equation here, so we subtract 81𝑎 from both sides. We divide both sides by negative 87. And swapping both sides and simplifying our fraction, we get that 𝑎 is equal to negative six over 29. That is the value of 𝑎 for which the function 𝑓 of 𝑥 is continuous at 𝑥 equals nine.