Video: Discussing the Differentiability of a Piecewise Function at a Point

Suppose 𝑓(π‘₯) = βˆ’1 + (3/π‘₯), if π‘₯ ≀ 1 and 𝑓(π‘₯) = βˆ’π‘₯Β³ + 3, if π‘₯ > 1. What can be said about the differentiability of the function at π‘₯ = 1?

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Video Transcript

Suppose that 𝑓 of π‘₯ is equal to negative one plus three over π‘₯ if π‘₯ is less than or equal to one and negative π‘₯ cubed plus three if π‘₯ is greater than one. What can be said about the differentiability of 𝑓 at π‘₯ equals one?

Here, we have been given a piecewise defined function and asked to check the differentiability. If we call our value of one π‘₯ nought, our general process for this type of question is first to check for continuity at π‘₯ nought. And then if this is satisfied, we check that the limit as π‘₯ approaches π‘₯ nought of our derivative exists, which means checking the left and right limits both exists and agree. We first begin with the continuity condition stated here. Since this is a piecewise defined function, we’ll need to check that both the left and the right limits as π‘₯ approaches one of 𝑓 of π‘₯ both exist and agree. When π‘₯ is less than one, our function is negative one plus three over π‘₯. By direct substitution, our left-sided limit then evaluates to two. When π‘₯ is greater than one, our function is negative π‘₯ cubed plus three. And by the same approach, the right-sided limit also evaluates to two.

Since these two limits both exist and agree, we can also see that the normal limit exists and is equal to two. Moving onto 𝑓 of one, the function itself here is defined by our first subfunction since π‘₯ is equal to one. And in fact, we’ve already substituted one into our first subfunction here, which gave us an answer of two. This means that we can say that the limit as π‘₯ approaches one of 𝑓 of π‘₯ and 𝑓 of one are both equal to two. We have, therefore, satisfied the continuity condition and our function is continuous when π‘₯ is equal to one. Let’s now move on to considering the derivative of our function.

Since our function is defind peacewise, the derivative of our function will also be defined piecewise. And we’ll need to differentiate both of our subfunctions. And here is worth noting that were not yet making any claims as the derivative when π‘₯ is equal to one, since that’s, in fact, what we’re trying to find right now. One of the tools that we can use to help us differentiate our subfunctions is the power rules stated here. To help us apply this rule more easily, we can reexpress three over π‘₯ as three π‘₯ to the power of minus one. Applying the rule, we find that 𝑓 dash of π‘₯ is equal to negative three over π‘₯ squared if π‘₯ is less than one and negative three π‘₯ squared if π‘₯ is greater than one.

We now need to consider the limit as π‘₯ approaches one of 𝑓 dash of π‘₯. We’ll do so by considering the left- and the right-sided limits in a process similar to the one that we just reformed for the continuity condition. By direct substitution of one, we find that both of these limits are equal to negative three. And since they exist and agree, the limit as π‘₯ approaches one of 𝑓 dash of π‘₯ is also equal to negative three. Since this limit exists, we know that our derivative also exists at π‘₯ equals one. We’ve now completed all the steps of our process. And from these, we can conclude that the function 𝑓 is differentiable at the point where π‘₯ is equal to one.

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