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### Course: Calculus, all content (2017 edition) > Unit 2

Lesson 8: Differentiability- Differentiability at a point: graphical
- Differentiability at a point: graphical
- Differentiability at a point: algebraic (function is differentiable)
- Differentiability at a point: algebraic (function isn't differentiable)
- Differentiability at a point: algebraic
- Differentiability at a point (old)

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# 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 about5:10Sal says that the function is defined for any x not equal to 1, then simplifies and evaluates it at 1.. Why is that correct?(25 votes)
- 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. :)(84 votes)

- In this case, what does differentiable and continuous mean? i'm still not getting it(6 votes)
- 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.(52 votes)

- At4:38, 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?(10 votes)
- 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.(13 votes)

- im not cut out for math(3 votes)
- Nobody is. Brains are not designed to be able to intuit these sort of deep mathematical truths. It is hard for everyone. The people who do it easily have spent hundreds or thousands of hours getting good at it, and they still make mistakes. My professor has a doctorate in pure mathematics and still makes mistakes. You gotta do the work, and the work sucks. But sucking at something is the first step towards being kinda good at something.(16 votes)

- At3:38where 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?(8 votes)
- 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).(5 votes)

- 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?(5 votes)
- 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.(4 votes)

- 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)?(5 votes)- 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.(4 votes)

- How can the slope be zero?(3 votes)
- 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.(5 votes)

- I thought a slope of 0 indicates a horizontal line. Why did Sal draw a vertical line for a slope equal to 0?(4 votes)
- 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 ?!(2 votes)

- Shouldn't lim as x-> 1- g(x) be equal to -2?(4 votes)
- For x < 1, for x → 1, we have g(x) → 0, because 0.9 - 1 = -0.1, 0.99 - 1 = -0.01, et cetera.

For x ≥ 1, for x → 1, we have g(x) = 0, because (1 - 1)² = 0.(2 votes)

## Video transcript

- [Voiceover] Is the function
given below continuous slash differentiable at x equals one? And they define the function g piece wise 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. And, like always, pause this video and see if you could figure this out. So let's do step by step. So first let's think about continuity. So for continuity, for g to
be continuous at x equals one that means that g of
one, that means g of one must be equal to the
limit as x approaches one of g of g of x. Well g of one, what is that going to be? G of one we're going
to fall into this case. One minus one squared 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 than 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, limit, and that's especially
useful 'cause we're in these different clauses
here as we approach from the left and the right hand side. So as x approaches one from the left hand side of g of x. Well we're going to be falling
into this situation here as we approach from the
left as x is less than one. So 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 could just substitute one in for x, and we get this is equal to zero. So 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 values are greater or equal to one it's gonna be x minus one squared. Well once again x minus one squared that is defined for all real numbers. It's continuous for all real numbers, so we could just pop that one in there. You get one minus one squared. Well that's just zero again, so the left hand limit,
the right hand limit are both equal 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. So we can rule out all of the ones that are saying that it's not continuous. So we can rule out that one, and we can rule out that
one right over there. So now let's think about
whether it is differentiable. So differentiability. So differentiability, I'll write
differentiability, ability. 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 one for f of x minus f of one 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 minus g of one over x minus one. And so let's just try
to evaluate this limit from the left and right hand sides, and we can simplify it. We already know that g of one 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 minus one or see if we can find the limit. So let's first think about the limit as we approach from the left hand side of g of x over x minus. G of x over x minus one. Well as we approach
from the left hand side, g of x is that right over there. So we could write this. Instead of writing g of x, we could write this as x minus one. X minus one over x minus one, and 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 minus one over x minus
one 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 of, once again, I could
write g of x of g of one, but g of one is just zero, so I'll just write g
of x over x minus one. Well what's g of x now? Well it's x minus one squared. So instead of writing g of x, I could write this as x minus one squared over x minus one, and so 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 minus one squared
divided by x minus one, well, that's just going
to give us x minus one. X minus one squared divided by x minus one is just going to be x minus one, and this limit, well this
expression right over here is going to be continuous
and defined for sure all x's that are not equaling one. Actually, let me, let me, well, it was before it was this, x minus one squared over x minus one. This thing over here, as
I said, is not defined for x equals one, but it
is defined for anything for x does not equal one, and
we're just approaching one. And, if we wanted to
simplify this expression, it would get, this would just be I think I just did this, but I'm making sure I'm doing it right. This is going to be the same this as that for x not being equal to one. Well this is just going to be zero. We could just evaluate when
x is equal to one here. This is going to be equal to zero. And 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 gonna look something like, we have a slope of one, so it's gonna look something like this. And then right when x is equal to one 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, continuous,
but not differentiable.