Maret School BC Calculus / Relationship between differentiability and continuity
Thus far in the course, the functions we have considered have been continuous. To say it in plain words, this means that we can draw the graph without lifting. For differentiability, that difference is required to converge even after being . that the limit from both sides of a value is equal to the function's value at that point. A function f(x) is continuous in the point x0 if the limit: limx→x0f(x). exists and is finite and equals the value of the function: f(x0)=limx→x0f(x).
So this is C comma F of C. So if you find the left side of the limit right over here, you're essentially saying okay, let's find this slope. And then let me get a little bit closer, and let's get X a little bit closer and then let's find this slope. And then let's get X even closer than that and find this slope.
And in all of those cases, it would be zero. The slope is zero. So one way to think about it, the derivative or this limit as we approach from the left, seems to be approaching zero. But what about if we were to take Xs to the right? So instead of our Xs being there, what if we were to take Xs right over here? If we get X to be even closer, let's say right over here, then this would be the slope of this line.
If we get even closer, then this expression would be the slope of this line. And so as we get closer and closer to X being equal to C, we see that our slope is actually approaching negative infinity. And most importantly, it's approaching a very different value from the right. This expression is approaching a very different value from the right as it is from the left. And so in this case, this limit up here won't exist. So we can clearly say this is not differentiable.
So once again, not a proof here. I'm just getting an intuition for if something isn't continuous, it's pretty clear, at least in this case, that it's not going to be differentiable. Let's look at another case. Let's look at a case where we have what's sometimes called a removable discontinuity or a point discontinuity. So once again, let's say we're approaching from the left. This is X, this is the point X comma F of X.
What is the difference between differentiability and continuity of a function?
Now what's interesting is where as this expression is the slope of the line connecting X comma F of X and C comma F of C, which is this point, not that point, remember we have this removable discontinuity right over here, and so this would be this expression is calculating the slope of that line. And then if X gets even closer to C, well, then we're gonna be calculating the slope of that line. If X gets even closer to C, we're gonna be calculating the slope of that line. And so as we approach from the left, as X approaches C from the left, we actually have a situation where this expression right over here is going to approach negative infinity.
And if we approach from the right, if we approach with Xs larger than C, well, this is our X comma F of X, so we have a positive slope and then as we get closer, it gets more positive, more positive approaches positive infinity.
But either way, it's not approaching a finite value. And one side is approaching positive infinity, and the other side is approaching negative infinity. This, the limit of this expression, is not going to exist.
So once again, I'm not doing a rigorous proof here, but try to construct a discontinuous function where you will be able to find this. It is very, very hard. And you might say, well, what about the situations where F is not even defined at C, which for sure you're not gonna be continuous if F is not defined at C.
Well if F is not defined at C, then this part of the expression wouldn't even make sense, so you definitely wouldn't be differentiable.
But now let's ask another thing. I've just given you good arguments for when you're not continuous, you're not going to be differentiable, but can we make another claim that if you are continuous, then you definitely will be differentiable?What is the Relationship between Differentiability and Continuity in Functions - Calculus Tips
Well, it turns out that there are for sure many functions, an infinite number of functions, that can be continuous at C, but not differentiable. So for example, this could be an absolute value function.
It doesn't have to be an absolute value function, but this could be Y is equal to the absolute value of X minus C. And why is this one not differentiable at C? Well, think about what's happening.
Think about this expression.
There are some situations which present us with a function which has an "unusual" point in fact, we'll see an example of this later on. Here is an example: Again, if we were to apply the methods we have from calculus to find the maxima or minima of this function, we would have to take this special point into consideration. Mathematicians have made an extensive study of discontinuities and found that they arise in many forms.
In practice, however, these are the principle types you are likely to encounter. Differentiability We have earlier seen functions which have points at which the function is not differentiable. An easy example is the absolute value function which is not differentiable at the origin. Notice that this function has a minimum value at the origin, yet we could not find this value as the critical point of the function since the derivative is not defined there remember that a critical point is a point where the derivative is defined and zero.
A similar example would be the function. Notice that which shows that the derivative does not exist at. However, this function has a minimum value at.
An example To illustrate how to deal with these kinds of situations, here is an example. Suppose that you are on one side of a lake listening to the radio. There is an announcement that you have won a special prize, but you must call the radio station quickly. The nearest phone is on the other side of the lake and you would like to reach the phone as quickly as possible. The situation is drawn to the left. The lake is shaped like a circle of radius 1 kilometer and there is a bridge running across the lake.
Continuity and Differentiability
You are at the point -1,0 while the phone is at the point. Let's find out how long it would take to reach the phone if you first swam across the lake at an angle t and then ran along the shore to the phone.
Remember that distance is equal to velocity times time so that the length of time it takes to swim is equal to the distance that you swim divided by the velocity at which you swim. To compute the distance you swimwe can use the law of cosines. It says that In other words, so that the time it takes you to swim is. Now, once you reach the opposite shore, you still have to run a distance along the shore. The phone is at a position which corresponds to a central angle of.