If f is continuous at a, must f be differentiable at a?

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Hi everyone, let's take a look at this practice problem. This problem says determine whether the given statement is true or false. And we're given the statement, if a function K is continuous at D, then K must be differentiable at D. We're given 4 possible choices as our answers. For choice A, we have true since continuity ensures a smooth curve. For choice B, false since a function can be continuous but have a sharp corner at D. For choice C, true because differentiability is a consequence of continuity. And for choice D, we have false because continuity only means the function has no breaks at D. So, we're asked to determine whether the statement is true or false, and we're gonna break the statement up into two parts. We're gonna look at the first half of this statement. And we see that we have that the A function K is continuous at D. So, what that means is that the limit as X purchase D of K of X. is equal to K of D. So this limit must be true if our function K is continuous at D. Now, for the second half of our statement, we see that K must be differentiable at D. And so, one way to define differentiability is that the limit As H approaches 0. Of the quantity of K of D+ H. Minus K of D. In quantity, divided by H, that this limit must exist. And so for the function to be differentiable at D then this limit must exist. Now, if we compare these two limit definitions, we see that the limit definition for a continuous function does not necessarily imply that the limit for um the derivative definition exists. And we're actually going to show a counter example of that. So we're gonna choose KFX. To be equal to the absolute value of X. I'm gonna set D equal to 0. So, we're gonna calculate both of these limits using this function. So, for the continuity limit, we will have the limit. As X approaches 0. Of the absolute value of X. And this is going to be equal to, and here I can just use direct substitution to evaluate this limit. And so that is equal to the absolute value of 0, which is equal to 0. So this limit does exist, and that's because the absolute value of X approaches 0 from both the left and right sides of X equal to 0, and its value is the same as being 0. So now we're going to look at the Um, derivative limit definition. So, here we're gonna be looking at the limit. As H Approaches 0. Of the quantity of the absolute value of D plus H, which in this case is D is equal to 0, so we'll have 0 plus H. Minus the absolute value of D, which is the absolute value of 0. And that whole quantity is divided by H. So, we can simplify this. This is going to be equal to the limit. As H approaches 0. Of the absolute value of H divided by H. And so now we need to see if this limit actually exists, because if we use direct substitution, we get 0 divided by 0, which is an indeterminate form. So we're going to look at the limit as H approaches 0 from the left and right sides. So let's start off with by approaching 0 from the right, so we'll have the limit. As H approaches 0 from the right. Of the absolute value of H divided by H. And this is going to be equal to the limit. As H approaches 0 from the right, in this case, H is going to be positive, since I'm to the right of 0, that means my values for H are going to be positive. And so that means the absolute value of H is just going to be equal to 8. So I'll have H divided by H. Which is equal to one. And so the limit as H approaches 0 from the right of 1 is just 1. So this whole limit evaluates to 1. We're gonna do the same thing, except we're going to approach 0 from the left. It will have the limit as H approaches 0 for the left of the absolute value of H divided by H. This is going to be equal to the limit. As H approaches 0 from the left, and here we have to be careful about the absolute value sign. So since we're approaching 0 from the left, that means all our values for H are negative. So the absolute value of H needs to be a positive quantity. So we can replace the absolute value of H with minus H. And that is still divided by H. And so, minus H divided by H is equal to -1. So, when we evaluate the limit of H protein zero from the left of -1, we get -1. And here we see that the two limits do not match. So that means that the limit As H approaches 0, since we get a different value whether we approach from the left or the right side of 0, that means that that limit of the absolute value of H divided by H. Does not exist or DNA. So, since our limit here does not exist, that means that the derivative does not exist at our point D or K. So, here we've shown that we have a function that is continuous at a point D, and the derivative does not exist. So that means that our statement here is actually false. And the reason it's false is because Even though our function is continuous, our derivative doesn't exist, and it doesn't exist because we actually have a sharp point at X equal to 0 for the absolute value of X. And so anytime we have a sharp point, our derivative is not defined. So the correct answer for this problem is actually answer choice B. That the statement is false, since the function can be continuous, but have a sharp corner at D, which makes the derivative not exist. So I hope that this explanation has been useful, and I'll see you in the next video.

If f is continuous at a, must f be differentiable at a?

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