Video: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios on an Interval

Discuss the continuity of the function 𝑓, given 𝑓(π‘₯) = sin (βˆ’8π‘₯ + 40)/(π‘₯ βˆ’ 5), if π‘₯ < 5 and 𝑓(π‘₯) = βˆ’8π‘₯Β²/25, if π‘₯ β‰₯ 5.

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Video Transcript

Discuss the continuity of the function 𝑓, given that 𝑓 of π‘₯ is equal to the sin of negative eight π‘₯ plus 40 divided by π‘₯ minus five if π‘₯ is less than five and 𝑓 of π‘₯ is equal to negative eight π‘₯ squared divided by 25 if π‘₯ is greater than or equal to five.

The question wants us to decide on which values of π‘₯ is the piecewise-defined function 𝑓 of π‘₯ continuous. And we can find the values of π‘₯ for which a piecewise-defined function 𝑓 of π‘₯ is continuous by following the following three steps. First, we want to find the domain of our piecewise-defined function 𝑓 of π‘₯ since a function can only be continuous at values of π‘₯ where it’s defined. Second, we want to check the continuity of our function 𝑓 of π‘₯ on each interval. Third, we just need to check that the endpoints of our intervals match. In other words, we’re just splitting the function 𝑓 of π‘₯ into its individual pieces and then checking at the end that they match up.

So, let’s start by finding the domain of our function 𝑓 of π‘₯. From the piecewise definition of our function 𝑓 of π‘₯, we can see when π‘₯ is less than five, our function 𝑓 of π‘₯ is equal to the sin of negative eight π‘₯ plus 40 divided by π‘₯ minus five. And this is the quotient of a trigonometric function and a polynomial, which are defined for all real values of π‘₯. So, this quotient will be defined as long as the denominator is not equal to zero. In other words, its domain is when π‘₯ is not equal to five.

However, we can see we only use values of π‘₯ less than five; π‘₯ is never equal to five. This tells us the sin of negative eight π‘₯ plus 40 divided by π‘₯ minus five is defined for all of these values of π‘₯. And in turn, this means our function 𝑓 of π‘₯ is defined for all values of π‘₯ less than five.

We can do the same for our second interval. When π‘₯ is greater than or equal to five, our function 𝑓 of π‘₯ is equal to negative eight π‘₯ squared divided by 25. And we can see that this is a polynomial, so it’s defined for all real values of π‘₯. So, we have negative eight π‘₯ squared divided by 25 is defined for all real values of π‘₯. In particular, it’s defined for all values of π‘₯ greater than or equal to five. So, we’ve shown 𝑓 of π‘₯ is defined for all values of π‘₯ less than five and for all values of π‘₯ greater than or equal to five. In other words, the domain of our function 𝑓 of π‘₯ is the entire set of real numbers.

So, we found the domain of our function 𝑓 of π‘₯. We now need to check the continuity of 𝑓 of π‘₯ on each of the intervals. And to check the continuity on each of these intervals, we recall the following three facts. Trigonometric functions are continuous on their domain, polynomials are continuous on the entire set of real numbers, and the quotient of two continuous functions is continuous on its domain.

Let’s start when π‘₯ is less than five. We see that our function 𝑓 of π‘₯ is equal to the sin of negative eight π‘₯ plus 40 divided by π‘₯ minus five. This is the quotient of a trigonometric function and a polynomial. So, by our rules, this function is continuous on its domain. And we already found its domain. Its domain is all the real values of π‘₯ except for five. So, if 𝑓 of π‘₯ is equal to the sin of negative eight π‘₯ plus 40 divided by π‘₯ minus five when π‘₯ is less than five and this is continuous, then 𝑓 of π‘₯ must also be continuous on this interval.

We can do something similar for our second interval, when π‘₯ is greater than or equal to five. We have that our function 𝑓 of π‘₯ is the polynomial negative eight π‘₯ squared divided by 25. And polynomials are continuous for all real values of π‘₯. So, when π‘₯ is greater than or equal to five, our function 𝑓 of π‘₯ is a polynomial, so it’s continuous on this interval. However, we have to be careful about when π‘₯ is equal to five. Remember, we need to check that the endpoints of our intervals match. So, in fact, this only shows us that our function 𝑓 of π‘₯ is continuous when π‘₯ is greater than five. We’ll need to check the case for π‘₯ is equal to five directly.

Let’s start with the endpoint of our interval when π‘₯ is less than five. Since our interval has π‘₯ strictly less than five, we’ll take the limit as π‘₯ approaches five from the left of the sin of negative eight π‘₯ plus 40 divided by π‘₯ minus five. This is the quotient of a trigonometric function and a linear function, so we can try using direct substitution. Substituting π‘₯ is equal to five, we get the sin of negative eight times five plus 40 divided by five minus five. And if we evaluate this expression, we get the sin of zero divided by zero, which is the indeterminate form zero divided by zero. So, we’re going to need to try and evaluate this limit in a different way.

To help us evaluate this limit, we recall the following standard limit law. The limit as π‘₯ approaches zero of the sin of π‘₯ divided by π‘₯ is equal to one. At first, it might seem difficult to see how we can use this result. However, we can notice that negative eight π‘₯ plus 40 is actually equal to negative eight times π‘₯ minus five.

And now, we can see something interesting; our limit has π‘₯ approaching five from the left. This means our denominator of π‘₯ minus five is approaching zero. And the factor of π‘₯ minus five in our numerator is also approaching zero. This is very similar to our limit law. This is motivation to rewrite our limit by using the substitution 𝑒 is equal to π‘₯ minus five.

Inside of our limit, by using this substitution, we get the sin of negative eight 𝑒 all divided by 𝑒. However, in our original limit, π‘₯ was approaching five from the left. In other words, π‘₯ is getting closer and closer to five while our values of π‘₯ are less than five. We can ask the question: what does this mean for our values of 𝑒? Well, 𝑒 is equal to π‘₯ minus five. And π‘₯ is getting closer and closer to five but always being less than five. This means 𝑒 is approaching zero and 𝑒 is less than zero. In other words, as π‘₯ approaches five from the left, 𝑒 approaches zero from the left.

So, now, we have the limit as 𝑒 approaches zero from the left of the sin of negative eight 𝑒 divided by 𝑒. And this is now very similar to our limit law, except we need to deal with the coefficient of negative π‘₯ in front of the π‘₯ inside of our sine function. We’ll start by rewriting all instances of π‘₯ with negative eight 𝑒. Next, we can take the constant factor of negative one-eighth outside of our limit. And if negative eight 𝑒 is approaching zero, this is the same as saying that 𝑒 is approaching zero.

Finally, we’ll just multiply through by negative eight, giving us the limit as 𝑒 approaches zero of the sin of negative eight 𝑒 divided by 𝑒 is equal to negative eight. And we can see this is exactly equal to what we have in our limit.

So, we found the first endpoint of our interval. It’s negative eight. It’s much easier to find the second endpoint of our interval. Since our interval is π‘₯ is greater than or equal to five, the endpoint will just be when π‘₯ is equal to five. So, we just substitute π‘₯ is equal to five into negative eight π‘₯ squared divided by 25. This gives us negative eight times five squared divided by 25, which we can evaluate to give us negative eight.

So, we can see, in both cases, we get negative eight. So, our function 𝑓 of π‘₯ is also continuous when π‘₯ is equal to five. So, we’ve shown our function 𝑓 of π‘₯ is continuous when π‘₯ is less than five, when π‘₯ is greater than five, and when π‘₯ is equal to five. Therefore, we’ve shown that the function 𝑓 is continuous for all real values of π‘₯.

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