Video: APCALC02AB-P1A-Q19-540138296031

Consider the function 𝑓(π‘₯) = (2π‘₯Β² βˆ’ π‘₯ βˆ’ 1)/(3(π‘₯ βˆ’ 1)) when π‘₯ β‰  1, and 𝑓(π‘₯) = π‘˜ when π‘₯ = 1. For what value of π‘˜ is 𝑓 continuous at π‘₯ = 1?

02:56

Video Transcript

Consider the function 𝑓 of π‘₯ is equal to two π‘₯ squared minus π‘₯ minus one all over three times π‘₯ minus one when π‘₯ is not equal to one, and π‘˜ when π‘₯ is equal to one. For what value of π‘˜ is 𝑓 continuous at π‘₯ equals one?

Here, we’ve got a piecewise function 𝑓 of π‘₯. We’ve been asked to find the value of π‘˜ so that the 𝑓 is continuous at the point π‘₯ equals one. Now, in order for some function 𝑔 to be continuous at some point π‘Ž, we require that the limit, as π‘₯ approaches π‘Ž, of 𝑔 of π‘₯ must be equal to 𝑔 of π‘Ž. Now, in our case, our function is 𝑓 of π‘₯. So we can call 𝑔 of π‘₯ 𝑓 of π‘₯. And we’re considering the continuity at π‘₯ is equal to one. So we can say that π‘Ž is equal to one.

So in order for our function to be continuous at π‘₯ is equal to one, we require that the limit as π‘₯ approaches one of 𝑓 of π‘₯ must be equal to 𝑓 of one. Looking at our function, we can find the value of 𝑓 of one. At π‘₯ is equal to one, 𝑓 is defined to be equal to π‘˜. Therefore, 𝑓 of one must equal π‘˜. Now, all we need to do is find the limit as π‘₯ approaches one of 𝑓 of π‘₯. Now, as π‘₯ approaches one, we can say that π‘₯ is not quite equal to one yet. Therefore, we have that π‘₯ is not equal to one. So we can say that 𝑓 of π‘₯ is equal to two π‘₯ squared minus π‘₯ minus one all over three π‘₯ minus one. And so, this is what we’re trying to find the limit of.

We can try to find this limit using direct substitution. However, we’ll quickly see that this gives us zero over zero. And this is undefined. However, since our function is a rational function with polynomials in the numerator and denominator, this does tell us, this does tell us that both polynomials will have a factor of π‘₯ minus one. So let’s factorise the polynomials in the numerator and denominator.

We see that our limit is equal to the limit, as π‘₯ approaches one, of two π‘₯ plus one multiplied by π‘₯ minus one all over three multiplied by π‘₯ minus one. We can see we have a factor of π‘₯ minus one in both the numerator and denominator. And so, we’re able to cancel them. However, we must be careful. Since in doing this, we’re in fact changing the function we’re taking the limit of. Since the domain of the original function did not include π‘₯ equals one. However, the domain of this function does. But we can, in fact, do this cancellation, since the limit of both functions will be equal.

And this will, in fact, help us to evaluate our limit. We get that this is equal to the limit, as π‘₯ approaches one, of two π‘₯ plus one over three. And, here, we can use direct substitution, which simplifies to give us three over three or one. Okay, so now we’ve evaluated the limit of our function, as π‘₯ tends to one. We can substitute this back into our equation. What we arrive at is that π‘˜ is equal to one. And this is, in fact, the value of π‘˜ for which 𝑓 is continuous at π‘₯ equals one.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.