Video: Finding the Value of the Variable That Make a Piecewise-Defined Function Continuous on Its Domains

Find the values of 𝑐 which make the function 𝑓 continuous at π‘₯ = 𝑐 if 𝑓(π‘₯) = 2 + π‘₯Β² if π‘₯ ≀ 𝑐, 𝑓(π‘₯) = βˆ’3π‘₯ if π‘₯ > 𝑐.

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Video Transcript

Find the values of 𝑐 which make the function 𝑓 continuous at π‘₯ equals 𝑐, if: 𝑓 of π‘₯ is equal to two plus π‘₯ squared, if π‘₯ is less than or equal to 𝑐; and negative three π‘₯, if π‘₯ is greater than 𝑐.

For a function 𝑓 of π‘₯ to be continuous at π‘₯ equals 𝑐, three things must be true. Firstly, 𝑓 of 𝑐 must exist, that is the function 𝑓 of π‘₯ is defined at π‘₯ equals 𝑐. Secondly, the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ must exist, so both the left-hand and right-hand limits exist and are equal. And thirdly, the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ must equal 𝑓 of 𝑐.

Let’s go through these one by one. First of all, we need 𝑓 of 𝑐 to exist. Well, when π‘₯ is less than or equal to 𝑐, which includes when π‘₯ is equal to 𝑐, 𝑓 of π‘₯ is equal to two plus π‘₯ squared. So 𝑓 of 𝑐 is equal to two plus 𝑐 squared. That is defined for any real value of 𝑐, so our first criterion is satisfied.

Now we move onto making sure that the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ exists. First, we evaluate the left-hand limit, the limit as π‘₯ tends to 𝑐 from below of 𝑓 of π‘₯. For this one-sided limit, we’re only considering values of π‘₯ for which π‘₯ is less than 𝑐, and so 𝑓 of π‘₯ is equal to two plus π‘₯ squared. And we evaluate this limit by direct substitution, replacing π‘₯ by 𝑐 to get two plus 𝑐 squared. Now we have to find the value of the other one-sided limit, the limit as π‘₯ tends to 𝑐 from above of 𝑓 of π‘₯. When π‘₯ is greater than 𝑐, 𝑓 of π‘₯ is equal to negative three π‘₯. So this is the limit as π‘₯ tends to 𝑐 of negative three π‘₯, which is negative three 𝑐. We have shown that both one-sided limits exist, but for the limit as π‘₯ tends to 𝑐 period of 𝑓 of π‘₯ to exist, the values of these two one-sided limits have to be equal. Two plus 𝑐 squared has to be equal to negative three 𝑐.

Let’s clear some room and solve this equation. First, we add three 𝑐 to both sides and rearrange the terms. Now we have a quadratic in 𝑐 in the form that we’re used to solving. We can factor this quadratic by inspection, and hence we see that these solutions are 𝑐 equals negative one and 𝑐 equals negative two. Of course, we could’ve used a different method, maybe completing the square or applying the quadratic formula. We would’ve got the same solutions. These are the two values of 𝑐, for which the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ will exist.

The only thing left to check is that the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ is equal to 𝑓 of 𝑐. Recall that two plus c squared was the limit as π‘₯ tends to 𝑐 from below of 𝑓 of π‘₯ and negative three 𝑐 was the limit as π‘₯ tends to 𝑐 from above of 𝑓 of π‘₯. We showed these two were equal when the 𝑐 is equal to negative one or negative two. And when they are equal, it makes sense to talk about the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯. Let’s first try 𝑐 equals negative one. In this case, the limit as π‘₯ tends to 𝑐 from below of 𝑓 of π‘₯ is two plus negative one squared by direct substitution, which is three. We also do direct substitution to find the limit as π‘₯ tends to 𝑐 from above, getting a negative three times negative one, which again is three. And as these two one-sided limits agree, we can say that the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ is three. To satisfy the third criterion, this limit must be equal to the value of the function at 𝑐. 𝑓 of 𝑐 is two plus 𝑐 squared, so when 𝑐 is negative one, 𝑓 of 𝑐 is two plus negative one squared, which is three. And we can see that the third criterion is satisfied, the limit as π‘₯ tends to 𝑐 of 𝑓 of 𝑐 is equal to 𝑓 of 𝑐 when 𝑐 is negative one.

Now we just have to check 𝑐 equals negative two. When 𝑐 is equal to negative two, as we know that the values of the two one-sided limits are equal, the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ is just the value of one of the one-sided limits. We’ll choose the limit as π‘₯ tends to 𝑐 from below of 𝑓 of π‘₯ which is two plus 𝑐 squared. Substituting the value of 𝑐, negative two in, we get two plus negative two squared, which is six. And of course, you can check this is exactly what we would get by direct substitution into the other one-sided limit. What’s the value of 𝑓 of 𝑐 in this case? Well, 𝑓 of 𝑐 is also equal to two plus 𝑐 squared. And substituting negative two, we get a value of six. And again, the third criterion is satisfied. The limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ is equal to 𝑓 of 𝑐; they’re both equal to six when 𝑐 is equal to negative two. And so in this case, the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ is equal to 𝑓 of 𝑐 whenever the limit as π‘₯ tends to 𝑐 of 𝑓 of π‘₯ exists, i.e., when 𝑐 is equal to negative one or 𝑐 is equal to negative two.

So there are two values of 𝑐 which make the function 𝑓 continuous at π‘₯ equals 𝑐, namely 𝑐 equals negative one and 𝑐 equals negative two.

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