Differentiability at a point: algebraic (function isn't differentiable) | Khan Academy

Is the function given below continuous differentiable at x equals 1? They define the function G piecewise right over here, and then they give us a bunch of choices: continuous but not differentiable, differentiable but not continuous, both continuous and differentiable, neither continuous nor differentiable. Like always, pause this video and see if you could figure this out.

So let's do this step by step. First, let's think about continuity. For G to be continuous at x equal 1, that means that G of 1 must be equal to the limit as x approaches 1 of G of x. Well, G of 1, what is that going to be? G of 1, we're going to fall into this case: 1 - 1^2 is going to be zero.

So if we can show that the limit of G of x as x approaches one is the same as G of one is equal to zero, then we know we're continuous there. Well, let's do the left and right-handed limits here. So if we do the left-handed limit, that's especially useful because we're in these different cases here.

As we approach from the left and the right-hand side, as x approaches one from the left-hand side of G of x, we're going to be falling into this situation here. As we approach from the left, as x is less than one, this is going to be the same thing as that. That's what G of x is equal to when we are less than one. As we're approaching from the left, well, this thing is defined, and it's continuous for all real numbers so we can just substitute one in for x and we get this is equal to zero.

So far, so good. Let's do one more of these. Let's approach from the right-hand side. As x approaches one from the right-hand side of G of x, well now we're falling into this case. So G of x, if we're to the right of one, if we're greater than or equal to one, it's going to be x - 1^2. Well, once again, x - 1^2 that is defined for all real numbers, is continuous for all real numbers so we could just pop that one in there. You get 1 - 1^2. Well, that's just zero again.

So the left-hand limit and the right-hand limit are both equal to zero, which means that the limit of G of x as x approaches one is equal to zero, which is the same thing as G of one. So we are good with continuity. We can rule out all the ones that say that it's not continuous, so we could rule out that one and we can rule out that one right over there.

Now let's think about whether it is differentiable. So differentiability—I'll write differentiability. Did I? Let's see, that's a long word. Differentiability! Alright, differentiability, what needs to be true here? Well, we have to have a defined limit as x approaches 1 for G of x minus G of 1 over... Oh, let me be careful, it's not F, it's G. So we need to have a defined limit for G of x - G of 1 over x - 1.

Let's just try to evaluate this limit from the left and right-hand sides and we could simplify. We already know that G of 1 is zero, so that's just going to be zero. So we just need to find the limit as x approaches one of G of x over x - 1, or see if we can find the limit.

Let's first think about the limit as we approach from the left-hand side of G of x over x - G of x over x - 1. Well, as we approach from the left-hand side, G of x is that right over there. Instead of writing G of x, we could write this as (x - 1)(x - 1) over (x - 1). As long as we aren't equal to one, this thing is going to be equal. As long as x does not equal one, (x - 1)/(x - 1) is just going to be one.

So this limit is going to be one. So that one worked out. Now let's think about the limit as x approaches one from the right-hand side. Once again, I could write G of x as G of 1, but G of 1 is just zero, so I'll just write G of x over x - 1. Well, what's G of x now? Well, it's (x - 1)^2.

So instead of writing G of x, I could write this as (x - 1)^2 over (x - 1). As long as x does not equal one, we're just doing the limit. We're saying as we approach one from the right-hand side. Well, this expression right over here, you have (x - 1)^2/(x - 1). Well, that's just going to give us (x - 1)(x - 1)/(x - 1).

So, as long as x does not equal one, this would just simplify to x - 1. This limit, well this expression right over here is going to be continuous and defined for sure for all x's that are not equaling one. Actually, let me... It was before this (x - 1)^2/(x - 1).

This thing right over here, as I said, it's not defined for x equals 1, but it is defined for anything not for x does not equal one, and we're just approaching one. If we wanted to simplify this expression, it would just be... I think I just did this, but I'm just making sure I'm doing it right. This is going to be equal to x - 1 as well.

So notice you get a different limit for this definition of the derivative as we approach from the left-hand side or the right-hand side, and that makes sense. This graph is going to look something like we have a slope of one so it's going to look something like this, and then right when x is equal to 1, and the value of our function is zero, it looks something like this.

It looks something like this, and so the graph is continuous. The graph for sure is continuous, but our slope coming into that point is one, and our slope right when we leave that point is zero. So it is not differentiable over there. So, it is continuous but not differentiable.