### Video Transcript

Find the values of π and π that make the function π continuous at π₯ is equal to negative two and π₯ is equal to two given that π of π₯ is equal to three π₯ minus five if π₯ is less than or equal to negative two and π of π₯ is equal to ππ₯ plus π if π₯ is greater than negative two and less than two and π of π₯ is equal to two π₯ squared minus three if π₯ is greater than or equal to two.

In this question, weβre given a piecewise-defined function π of π₯, which contains two unknowns, π and π. We need to determine the values of these unknowns π and π, which will make π continuous at π₯ is equal to negative two and π₯ is equal to two. And there are many different ways of answering this question. For example, we can actually answer this question graphically. We could sketch the first and second subfunction over their subdomains. And then we would note the second subfunction of π of π₯ is a linear function. And both π₯ is equal to negative two and π₯ is equal to two are the endpoints of its subdomain.

So we could note for our function to be continuous at negative two and at two, our linear function would need to connect the endpoint of the first subfunction to the endpoint of the third subfunction. However, in this video, weβre going to take a more rigorous approach from the definition of continuity. We recall that we say a function π is continuous at π₯ is equal to π if the following three conditions hold. First, π evaluated at π needs to be defined. In other words, π needs to be in the domain of our function π.

Next, we need the limit as π₯ approaches π of π of π₯ to exist. Finally, the limit as π₯ approaches π of π of π₯ needs to be equal to the function evaluated at π. And we need all three conditions to hold. In fact, since we need our function to be continuous at negative two and at two, we need all three conditions to hold for two different values of π. Letβs start by making our function continuous at π₯ is equal to negative two.

First, we need to check that π evaluated at negative two is defined. We can do this directly from the piecewise definition of our function. We can see negative two is in the first subdomain, so our function is defined at π₯ is equal to negative two. However, itβs worth noting we need π evaluated at negative two for our third condition. So we might as well evaluate this anyway. Since π₯ is equal to negative two is in the first subdomain, we substitute negative two into the first subfunction.

π evaluated at negative two is three times negative two minus five which is equal to negative 11. We now want to check the second continuity condition when π₯ is negative two. We need to check the limit as π₯ approaches negative two of π of π₯ exists. And weβll do this by noting negative two is one of the endpoints of our subdomains. So weβre going to do this by checking its left and right limits are equal. Letβs start with the limit as π₯ approaches negative two from the left of π of π₯. Since weβre taking the limit as π₯ approaches negative to from the left, our values of π₯ will be less than negative two. And we can see when π₯ is less than negative two, our function is equal to three π₯ minus five, which means their limits as π₯ approaches negative two from the left will be equal.

In particular, since this is a linear function, we can evaluate this by direct substitution. And this is, of course, the same as π evaluated at negative two. We get three times negative two minus five, which is equal to negative 11. We can now follow the same process to evaluate the limit as π₯ approaches negative two from the right of π of π₯. First, since π₯ is approaching negative two from the right, our values of π₯ are greater than negative two. Weβll write this as negative two is less than π₯. And this is because the second subdomain of our function is π₯ is greater than negative two and less than two.

Remember, when weβre taking the limit of a function, we want our values of π₯ to get closer and closer to the limit point. So in particular, we can choose our values of π₯ less than two. And this means this limit must be equal to the limit as π₯ approaches negative two from the right of the second subfunction, ππ₯ plus π. And this is once again a linear function, so we can evaluate this limit by direct substitution. And if we substitute π₯ is equal to negative two, we get negative two π plus π.

But remember, weβre trying to show that the limit as π₯ approaches negative two of our function π of π₯ exists. And for this limit to exist, we need the left and right limit as π₯ approaches negative two of π of π₯ to be equal. So we need negative two π plus π is equal to negative 11 for the second continuity condition to hold true. This then sets the limit as π₯ approaches negative two of our function π of π₯ to be equal to negative 11, which we can see is equal to π evaluated at negative two. So all three continuity conditions will hold for our function π at π₯ is equal to negative two, provided negative two π plus π is equal to negative 11. But this is a linear equation in two unknowns, so we canβt solve this for π and π.

So instead, letβs clear some space and use the fact that our function needs to be continuous at π₯ is equal to two. We can apply the same process in our definition. First, we need our function to be defined at π₯ is equal to two. Once again, we can see this directly from the definition of π of π₯. Two is in the third subdomain of our function, so we substitute π₯ is equal to two into π of π₯ to get that π evaluated at two is two times two squared minus three, which is five. So the first condition for continuity at π₯ is equal to two holds true.

Weβll follow the same process to check the limit. Weβll check the left and right limit as π₯ approaches two. Letβs start with the limit as π₯ approaches two from the right of π of π₯. Our values of π₯ are greater than two. And this means our function π of π₯ is equal to its third subfunction. So this is equal to the limit as π₯ approaches two from the right of two π₯ squared minus three. And this is a polynomial, so we can evaluate this limit by using direct substitution. We get two times two squared minus three, which is equal to five.

Letβs now evaluate the limit as π₯ approaches two from the left of π of π₯. Since π₯ is approaching two, we can restrict our values of π₯ to be greater than negative two and less than two. And for these values of π₯, our function π of π₯ is equal to ππ₯ plus π. Therefore, we just need to evaluate the limit as π₯ approaches two from the left of ππ₯ plus π. And this is a linear function, so we can do this by using direct substitution. We substitute in π₯ is equal to two, and we get to π plus π.

And now, since we want our function π of π₯ to be continuous when π₯ is equal to two, we need the limit as π₯ approaches two from the right of π of π₯ to be equal to the limit as π₯ approaches two from the left of π of π₯. We need two π plus π to be equal to five. And before we solve these simultaneous equations, itβs worth noting if two π plus π is equal to five, then the left and right limit as π₯ approaches two of π of π₯ are equal. So the second continuity condition holds true. And five is also equal to π evaluated at π. So the limit as π₯ approaches two of π of π₯ will be equal to five, which is equal to π evaluated at π.

Therefore, what weβve shown is if our values of π and π satisfy the first equation, then our function π is continuous at π₯ is equal to negative two. And if our values of π and π satisfy the second equation, then our function is continuous at π₯ is equal to two. So we just need to find values of π and π, which satisfy both equations.

And weβll do this by adding the two equations together. Negative two π plus two π is equal to zero and π plus π is equal to two π. So the left-hand side of our equation is two π, and negative 11 plus five is negative six. So we get two π is equal to negative six. We then divide the equation through by two to see that π is equal to negative three.

We can then solve for π by substituting π is equal to negative three into either of our simultaneous equations. Weβll substitute this into our first simultaneous equation. We get negative two π minus three is equal to negative 11. And we can then solve this equation for π. We add three to both sides of the equation to get negative two π is equal to negative eight. And then we divide through by negative two. We get π is equal to four, which gives us our final answer.

Therefore, we were able to show for our function π to be continuous at π₯ is equal to negative two and π₯ is equal to two, the value of π needs to be equal to four and the value of π needs to be equal to negative three.