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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 2

Lesson 4: Connecting differentiability and continuity: determining when derivatives do and do not exist

# Differentiability at a point: algebraic (function isn't differentiable)

We examine a piecewise function to determine its differentiability and continuity at an edge point. Through analyzing left and right-hand limits, we find that the function is continuous at the point. However, due to differing slopes from the left and right, the function is not differentiable at the edge point.

## Want to join the conversation?

• At about Sal says that the function is defined for any x not equal to 1, then simplifies and evaluates it at 1.. Why is that correct?
• Good question! This is a tricky concept in limits, but also a very common and important one. Here is the KEY POINT: when we are trying to evaluate a limit as x → 1, what we are precisely interested in is what value the function APPROACHES, as x APPROACHES 1. That is, what does its value head toward when x gets VERY CLOSE to 1. Well, the first thing we always try is to simply plug in x = 1, because unless the function is something quite strange it will APPROACH the value it actually EQUALS when x actually equals 1.

Now, very often, we will find that the expression happens to be undefined at EXACTLY x = 1, usually because it equals 0/0 when we plug in x = 1. But that often means that both the numerator and the denominator have a factor of (x - 1), or can be made to have a factor of (x - 1) after a bit of work. So those two factors of (x - 1), top and bottom, can be cancelled for any value of x EXCEPT when x equals EXACTLY 1. But the idea of limits is that while we are approaching extremely close to 1, we don't let the true value of x be exactly 1. Maybe x = 0.99999999999999, or x = 1.0000000000001, but not exactly 1. Thus, we can cancel the factors of (x-1), since x is close to, but not equal to 1.

Now, we have a new, simpler expression, which is exactly equal to the original expression (we have simply cancelled (x-1)/(x-1) ), and we still want to know what it APPROACHES as x APPROACHES 1, so we try plugging in x = 1...

I know at first it might seem like a self-contradictory process, but if you think about it, it's not. That last sentence above, "plugging in x = 1..." might make you yell, "But you just said x cannot be exactly 1!" The thing is, you have to see "plugging in x = 1" as more like a "let's just suppose x could be 1 and see what would happen", and not "we are now making x officially equal to 1". Limits are a slippery and tricky idea, because they are sneakily based on a sort of idea of infinitely tiny increments, and dealing mathematically with either the infinitely large or the infinitely small always takes some getting used to. :)
• In this case, what does differentiable and continuous mean? i'm still not getting it
• Continuous means that you can trace the line with a pencil without picking up the pencil from the paper. There's no gaps, jumps, holes or any of that in the line; just one long line without taking the pencil of the paper. One caveat to that: sharp turns.

Differentiable means you can find a derivative at that point, i.e., you can find the slope there. BUT, the slope as you approach from the left MUST BE the same slope as the slope coming from the right to that same point. With a sharp turn, these two slopes are different, and thus, you can't find a derivative there.
• At , Sal explains that (x-1)^2/ (x-1) is equal to x-1 which makes sense, but couldn't this also be solved as a difference of squares in which the numerator is factored into (x+1) and (x-1)? Then the (x-1) in the numerator and denominator would cancel out to leave x+1 which equals 2? Is this mathematically correct? Is it possible to have different answers for this one-sided limit that still prove that the derivative doesn't exist because it's not the same as the other one?
• The numerator is not a difference of squares and does not factor into (x+1)·(x-1).
(x-1)² = (x-1)·(x-1) = x² - 2x + 1.
• At where does that formula where he's finding the lim as x approaches 1(from left) of g(x)/(x-1) come from, specifically the x-1 on the bottom?
• From the definition of the derivative. The definition is the limit as x approaches 1 of (g(x) - g(1)) / (x - 1). But g(1) is 0, so we are left with g(x) / (x - 1).
• If a function is non-differentiable at a point, then how can you create a formula that will give you the slope of the tangent line at any point of the function? Can you at all?
• Yes, you can define the derivative at any point of the function in a piecewise manner. If f(x) is not differentiable at x₀, then you can find f'(x) for x < x₀ (the left piece) and f'(x) for x > x₀ (the right piece). f'(x) is not defined at x = x₀.

For example, f(x) = |x - 3| is defined and continuous for all real numbers x. It is differentiable for all x < 3 or x > 3, but not differentiable at x = 3. Here is our piecewise derivative: for all x < 3, f'(x) = -1; for all x > 3, f'(x) = 1; and at x = 3, f'(x) is undefined.
• Instead of using limits to show the function is not differentiable, could we just work out the derivatives for each part and say the function is non-differentiable because they are different at x=1?

What I mean is, for x<1 the derivative would be d/dx(x-1)= 1, and for x>1 it would be d/dx(x-1)^2=2(x-1). So evaluating at x=1, the derivative for the first part would be 1 and for the second part it would be 2(1-1)=0 (which correspond to the limits that Sal worked out). This seems to imply that the function is not differentiable at x, right?

And if so, could we go the other way and show that the function is differentiable by computing the derivative for each part and showing the derivatives are equal at the point where the two parts meet (x=1, in this case)?
• Hi Jonathan,
Are you referring to a piece-wise defined function? If so, then yes. However, you will have to determine that the function is continuous at the point in question as well. There could be a piece-wise function that is NOT continuous at a point, but whose derivative implies that it is. So if a function is piece-wise defined and continuous at the point where they "meet," then you can create a piece-wise defined derivative of that function and test the left and right hand derivatives at that point.
• How can the slope be zero?
• That will be the case when the slope is neither increasing or decreasing, i.e. if the slope is a horizontal line. Slope is defined as rise over run. When there is no rise or fall to the slope, then vertically nothing happens, right? I.e., rise over run is y / x, with y being 0.
• I thought a slope of 0 indicates a horizontal line. Why did Sal draw a vertical line for a slope equal to 0?
• Even I confused myself at this point of the video. If you notice he didn't draw vertical line but
1) a slanting line to represent negative side of 1 i.e x<1 which is a linear progression and represented by y=g(x)==x-1
2) a curve to present positive side of 1 i.e x >1 (ignoring x=1 for now) which is a non-linear progression and represented by y=g(x)=(x- 1)²
The point x=1 is still represented by y=g(x) = (x - 1)² because of condition x ≥1 for (x - 1)² in g(x) function definition

So he didnt draw any line to represent slope

Yes you are correct, only horizontal line will have slope 0. Now to understand where this horizontal line is coming, let us recollect definition of Derivative : is slope of tangent line at given point (in our case it is x=1). If you try to draw a tangent line at x=1 in such way it is touching only x=1 but not any other point on Linear line (x<1) nor on Non-Linear Curve(x>1), it would just pass thru the point x =1 i.e Horizontal Line with the slope - 0. Hope this helps ?!
• Shouldn't lim as x-> 1- g(x) be equal to -2?