Differentiability at a point: algebraic (function is differentiable) | AP Calculus AB | Khan Academy

Is the function given below continuous differentiable at x = 3? And they've defined it piecewise, and we have some choices: continuous, not differentiable, differentiable, not continuous, both continuous and differentiable, neither continuous nor differentiable.

Now, one of these we can knock out right from the get-go: in order to be differentiable, you need to be continuous there. So, you cannot have differentiable but not continuous, so let's just rule that one out.

Now, let's think about continuity. If it isn't continuous, then it's not going to be differentiable. So, let's think about it a little bit. In order to be continuous, f(3) needs to be equal to the limit of f(x) as x approaches 3. Now, what is f(3)? Well, let's see. We fall into this case right over here because x is equal to 3. So, 6 * 3 is 18, and 18 - 9 is 9, so this is 9.

The limit of f(x) as x approaches 3 needs to be equal to 9. Let's first think about the limit as we approach from the left-hand side. The limit as x approaches 3 from the left-hand side of f(x): well, when x is less than 3, we fall into this case, so f(x) is just going to be equal to x². This is defined and continuous for all real numbers, so we can just substitute the three in there. This is going to be equal to 9.

Now, what's the limit as we approach 3 from the right-hand side of f(x)? Well, as we approach from the right, this one right over here is f(x) = 6x - 9. So, we just write 6x - 9. Once again, 6x - 9 is defined and continuous for all real numbers, so we can just pop a three in there, and you get 18 - 9. Well, this is also equal to 9.

So, the left-hand and right-hand limits both equal 9, which is equal to the value of the function there. So it is definitely continuous. Thus, we can rule out this choice right over there.

Now let's think about differentiability. In order to be differentiable, the limit as x approaches 3 of (f(x) - f(3)) / (x - 3) needs to exist. So, let's see if we can evaluate this. First of all, we know what f(3) is. We have already evaluated this. This is going to be 9.

Let's see what the limit is as we approach from the left-hand side and the right-hand side, and if they're approaching the same thing, then we know that the same thing they're approaching is the limit.

So let's first think about the limit as x approaches 3 from the left-hand side. It’s (f(x) - 9) / (x - 3). But as we approach from the left-hand side, f(x) as x is less than 3 is equal to x². So, instead of f(x) - 9, I'll write x² - 9.

Now, x² - 9 is a difference of squares. So this is (x + 3)(x - 3). These would cancel out, and we can say that this is equivalent to x + 3 as long as x does not equal 3. That's okay because we're approaching from the left. As we approach from the left, well, x + 3 is defined for all real numbers, it's continuous for all real numbers, so we can just substitute the three in there. We would get a 6.

Now let's try to evaluate the limit as we approach from the right-hand side. Once again, it's f(x), but as we approach from the right-hand side, f(x) is 6x - 9, that’s our f(x), and then we have minus f(3), which is 9. So it’s 6x - 18.

6x - 18, well, that's the same thing as 6(x - 3). As we approach from the right, well, that's just going to be equal to 6.

So it looks like our derivative exists there, and it is equal to the limit as x approaches 3 of all of this business equals 6 because the limit as we approach from the left and the right is also equal to 6.

So this looks like we are both continuous and differentiable.