# Video: Finding the Values of Variables That Make a Piecewise-Defined Function Continuous at Two Points

Find the values of π and π that make the function π continuous at π₯ = β1 and π₯ = β6, given that π(π₯) = 3π₯ + 11 if π₯ β€ β6, π(π₯) = ππ₯ + π if β6< π₯< β1, π(π₯) = β5π₯Β² + 10 if π₯ β₯ β1.

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### Video Transcript

Find the values of π and π that make the function π continuous at π₯ equals negative one and π₯ equals negative six, given that π of π₯ is equal to three π₯ plus 11 if π₯ is less than or equal to negative six, ππ₯ plus π if π₯ is between negative six and negative one, and negative five π₯ squared plus 10 if π₯ is greater than or equal to negative one.

We know exactly what the function does in the regions π₯ is less than or equal to negative six and π₯ is greater than or equal to negative one. What we donβt know is what precisely happened when π₯ is between negative six and negative one. Weβre given an expression for the function in this region in terms of π and π. And our task is to find the values of π and π in a way that makes the function continuous at the endpoints of this region π₯ equals negative one and π₯ equals negative six.

It might be a good idea to represent what we do know on a graph, so we can see what weβre doing. On the left-hand side of the graph, when π₯ is less than or equal to negative six, we have a bit of the line with equation π¦ equals three π₯ plus 11. And on the right side of the graph, when π₯ is greater than or equal to negative one, we have a bit of the parabola with equation π¦ equals negative five π₯ squared plus 10.

What we want to know is what happens when π₯ is between negative six and negative one. Between these two points, π of π₯ is equal to ππ₯ plus π, which we can recognize as the equation of a line. And to make this function π continuous, weβre going to have to join up the dots with the other two parts of the graph. Stop! This must be what the graph of our function looks like. Letβs see if we can use it to help us find the values of π and π.

To make our function continuous at π₯ equals negative six, we had to draw a line starting at this point circled. What is this point? Its π₯-coordinate is negative six and itβs on the graph of our function π, so its coordinates must be negative six, π of negative six. And when π₯ is negative six, weβre in this region here when π of π₯ is equal to three π₯ plus 11. So we substitute negative six into this expression and find that π of negative six is negative seven.

We know that the line segment that weβve drawn with equation π¦ equals π π₯ plus π must go through this point. And substituting these coordinates in, we get that negative seven is equal to π times negative six plus π. And swapping things around a bit, we get negative six π plus π equals negative seven. This is a condition on π and π for π of π₯ to be continuous at π₯ equals negative six.

And we can see how this relates to the more formal definition of continuity. On the left, we have negative seven, which came from π of negative six and on the right-hand side, we have the limit as π₯ tends to negative six from above of π of π₯. These two things being equal to each other is one of the things that we require for continuity at this point.

The other thing that we require is that the limit as π₯ tends to negative six from below of π of π₯ is also equal to these two things, but we get this for free when π₯ is less than or equal to negative six π of π₯ is given by a nice continuous rule. And so the limit as π₯ tends to negative six from below of π of π₯ is equal to π of negative six. Anyway, weβve seen that to make π continuous at π₯ equals negative six, we need negative six π plus π to be equal to negative seven.

Letβs clear some space and see what the equivalent condition is for continuity at π₯ equals negative one. To do this, we find the coordinates of the other endpoint of our line segment. The π₯-coordinate is negative one. And of course, because this point lies on the graph the π¦-coordinate must be π of negative one. What is π of negative one? We substitute into the expression for π of π₯ when π₯ is greater than or equal to negative one, and we find that itβs five. And because π of π₯ has a nice continuous rule in the region π₯ is greater than or equal to negative one, this value of five is also the limit as π₯ tends to negative one from above of π of π₯.

Now, whatβs the limit as π₯ tends to negative one from below of π of π₯? Well, now, weβre in the region where π₯ is between negative six and negative one. And we just substitute in π₯ is negative one to this expression to get π times negative one plus π, which is minus π plus π. And we want all these three things to be equal to satisfy our definition of continuity at the point π₯ equals negative one. So minus π plus π must be equal to five. This is of course exactly the same condition we would get by requiring that the line segment has to pass through the point negative one, five, which was the endpoint of our bit of parabola.

We now have two conditions on π and π: that negative π plus π must be equal to five for continuity at π₯ equals negative one and the condition that negative six π plus π equals negative seven, which came from the requirement for continuity at π₯ equals negative six. So to find the values of π and π for which π is continuous at both π₯ equals negative one and π₯ equals negative six, we need to solve these two linear equations in π and π simultaneously.

There are many methods for doing this. Iβm going to subtract the second equation from the first to get that negative five π is equal to negative 12. And hence, that π is equal to negative 12 over negative five, which is just 12 over five. And we can substitute this value of π into one of our equations to get that negative 12 over five plus π must be equal to five. And hence, that π is equal to 37 over five. So there we have it. The values of π and π that make the function π continuous at π₯ equals negative one and negative six are π equals 12 over five and π equals 37 over five.