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

Discuss the differentiability of the function 𝑓 at π‘₯ = 1 given 𝑓(π‘₯) = 2π‘₯ + 8, if π‘₯ < 1 and 𝑓(π‘₯) = π‘₯Β² + 9, if π‘₯ β‰₯ 1.

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Video Transcript

Discuss the differentiability of the function 𝑓 at π‘₯ is equal to one given that 𝑓 of π‘₯ is equal to two π‘₯ plus eight if π‘₯ is less than one and 𝑓 of π‘₯ is equal to π‘₯ squared plus nine if π‘₯ is greater than or equal to one.

We’re given a piecewise-defined function 𝑓 of π‘₯. We need to determine whether the function 𝑓 is differentiable when π‘₯ is equal to one. There’s a few different ways of doing this. For example, we could first check whether our function 𝑓 of π‘₯ is continuous when π‘₯ is equal to one. Then if 𝑓 of π‘₯ is continuous when π‘₯ is equal to one, we can check the slope from the left and from the right when π‘₯ is equal to one by differentiating both pieces of our function. If these two slopes match, then we can say 𝑓 of π‘₯ is continuous when π‘₯ is equal to one.

However, in this video, we’re going to determine this directly from the definition of a derivative. Let’s start by recalling what we mean by the derivative of 𝑓 of π‘₯ with respect to π‘₯ at a point π‘₯ naught. This is equal to the limit as β„Ž approaches zero of 𝑓 of π‘₯ naught plus β„Ž minus 𝑓 of π‘₯ naught all divided by β„Ž. And this assumes this limit exists. If this limit does not exist, then we say that our function 𝑓 of π‘₯ is not differentiable at the point π‘₯ naught. In this case, we’re given the piecewise function 𝑓 of π‘₯. And we want to know if it’s differentiable at the point where π‘₯ is equal to one. So we’ll set π‘₯ naught equal to one.

We can then substitute π‘₯ naught is equal to one into our definition of the derivative. So determining whether 𝑓 of π‘₯ is differentiable when π‘₯ is equal to one is the same is determining whether this limit exists. However, we can’t directly evaluate this limit because our function 𝑓 of π‘₯ is given as a piecewise function and π‘₯ one is the endpoints of one of these intervals.

So we need to remember a fact about limits. Instead of finding the limit as β„Ž approaches zero of this expression, we’ll find the limit as β„Ž approaches zero from the right of this expression and the limit as β„Ž approaches zero from the left of this expression. If both of these limits exist and are equal, then we know our function 𝑓 of π‘₯ is differentiable at this point. However, if either limit does not exist or these limits are not equal, then we know that our function is not differentiable when π‘₯ is equal to one.

Let’s start by evaluating the limit as β„Ž approaches zero from the left of 𝑓 of one plus β„Ž minus 𝑓 of one all divided by β„Ž. To evaluate this limit, we first need to notice that β„Ž is approaching zero from the left. This means that all of our values of β„Ž will be less than zero. And if β„Ž is less than zero, then one plus β„Ž will be less than one. And we know from our piecewise definition of the function 𝑓 of π‘₯ when π‘₯ is less than one, 𝑓 of π‘₯ is equal to the linear function two π‘₯ plus eight. So in this case, to evaluate 𝑓 at one plus β„Ž, we just substitute π‘₯ is equal to one plus β„Ž into the linear function two π‘₯ plus eight. This gives us two times one plus β„Ž add eight.

We might be tempted to substitute one into this linear function. But remember, we need to use the piecewise definition of 𝑓 of π‘₯. When π‘₯ is equal to one, 𝑓 of π‘₯ is exactly equal to the quadratic π‘₯ squared plus nine. Substituting π‘₯ is equal to one into our quadratic gives us one squared plus nine. So we now have to evaluate the limit as β„Ž approaches zero from the left of two times one plus β„Ž plus eight minus one squared plus nine all divided by β„Ž.

To do this, we need to simplify the expression inside of our limit. We’ll start with the first term in our numerator. We’ll distribute two over the parentheses. This gives us two plus two β„Ž plus eight. And of course, we can simplify this since two plus eight is equal to 10. Next, we’ll evaluate the second term in our numerator. we have one squared plus nine is equal to one plus nine. And of course, we can simplify one plus nine to give us 10.

So now, we have the limit as β„Ž approached zero from the left of two β„Ž plus 10 minus 10 all divided by β„Ž. Of course, we can keep simplifying this. In our numerator, 10 minus 10 is equal to zero. So in actual fact, this entire limit just simplified to give us the limit as β„Ž approaches zero from the left of two β„Ž divided by β„Ž. And of course, we know β„Ž is approaching zero from the left. In particular, this means that β„Ž is not equal to zero. It’s just getting closer and closer to zero. So we can cancel the shared factor of β„Ž in our numerator and our denominator, meaning we’re just left with the limit as β„Ž approaches zero from the left of the constant two. But two is a constant, so this limit evaluates to give us two.

Therefore, as π‘₯ approaches one from the left, the slope of our function 𝑓 of π‘₯ is equal to two. We now need to do exactly the same thing from the right. So let’s clear some space and evaluate the limit as β„Ž approaches zero from the right of 𝑓 of one plus β„Ž minus 𝑓 of one all divided by β„Ž. This time, we have β„Ž is approaching zero from the right. This means we know all of our values of β„Ž will be bigger than zero. And if β„Ž is just bigger than zero, this means one plus β„Ž will be bigger than one.

So once again, we need to use the piecewise definition of our function 𝑓 of π‘₯ to evaluate 𝑓 of one plus β„Ž. We know when π‘₯ is greater than or equal to one, 𝑓 of π‘₯ is exactly equal to the quadratic π‘₯ squared plus nine. So because one plus β„Ž is greater than one, to evaluate 𝑓 at one plus β„Ž, we substitute π‘₯ is equal to one plus β„Ž into this quadratic. This gives us one plus β„Ž all squared plus nine. We already found 𝑓 evaluated at one in our previous limit. To do this, we use the piecewise definition of 𝑓 of π‘₯.

We just substitute π‘₯ is equal to one into the quadratic π‘₯ squared plus nine. This gave us one squared plus nine, which we simplified to give us 10. So we now need to evaluate the limit as β„Ž approaches zero from the right of one plus β„Ž all squared plus nine minus 10 all divided by β„Ž. To do this, we can start simplifying. First, we know that nine minus 10 is equal to negative one. Next, we want to distribute the square over our parentheses.

Doing this by using the FOIL method or by using binomial expansion, we get one plus two β„Ž plus β„Ž squared. Then we need to subtract one from this and divide through by β„Ž. And in fact, we can see we can keep simplifying. In our numerator, we have one minus one, which simplifies to give us zero. So this gives us the limit as β„Ž approaches zero from the right of two β„Ž plus β„Ž squared all divided by β„Ž. And once again, this limit is as β„Ž is approaching zero from the right. β„Ž is getting closer and closer to zero, but β„Ž is never equal to zero.

This means we can cancel the shared factor of β„Ž in our numerator and our denominator. This leaves us with the limit as β„Ž approaches zero from the right of two plus β„Ž. And of course, as β„Ž is approaching zero from the right, our value of β„Ž is approaching zero and our constant two remains constant. So this limit just evaluates to give us two. And now, we can see directly from the definition of the slope that these two slopes are equal. And because these two values are equal, we can conclude that 𝑓 of π‘₯ is differentiable at π‘₯ is equal to one.

It’s also worth reiterating at this point, we could only use this method because we directly worked with the definition of differentiability at a point. If we had instead wanted to work with other rules of differentiation, such as the power rule for differentiation on our function 𝑓 of π‘₯, we would also need to show that 𝑓 of π‘₯ is continuous when π‘₯ is equal to one. However, as we’ve seen, it’s not necessary to use this method. We can just do this directly from the definition of a derivative. Therefore, we were able to show the function 𝑓 of π‘₯ is equal to two π‘₯ plus eight if π‘₯ is less than one and 𝑓 of π‘₯ is equal to π‘₯ squared plus nine if π‘₯ is greater than or equal to one is differentiable at π‘₯ is equal to one.

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