Review your knowledge of the mean value theorem and use it to solve problems.
Log in meghna rana 7 years agoPosted 7 years ago. Direct link to meghna rana's post “How is mean value theorem...” How is mean value theorem applied in real life? • (12 votes) jude4A 7 years agoPosted 7 years ago. Direct link to jude4A's post “Mean value theorem can be...” Mean value theorem can be used to determine the speed of something, like a policeman using a speedometer. (33 votes) Lori Feng 8 years agoPosted 8 years ago. Direct link to Lori Feng's post “In the question, the inte...” In the question, the interval is a closed one, but in the explanation, the interval is open, thus excluding 3 as an answer. Why does the explanation have an open interval? • (7 votes) amal.bukhari 8 years agoPosted 8 years ago. Direct link to amal.bukhari's post “A and B are given in a cl...” A and B are given in a closed interval in the question because the hypotheses of the mean value theorem include that the function must be continuous on the interval [A,B] and differentiable on the interval (A,B). The function does not have to be differentiable at A and B because you are looking for a value c in between A and B that is equal to the slope of the secant line connecting A and B. Hope that helps. (15 votes) Urikhan 9 months agoPosted 9 months ago. Direct link to Urikhan's post “Why c would be only in op...” Why c would be only in open interval? Because a and b are differentiable over open interval? • (1 vote) Venkata 9 months agoPosted 9 months ago. Direct link to Venkata's post “Correct. We do need to fi...” Correct. We do need to find f'(c) after all. So, c needs to be a point where the derivative exists (i.e. a differentiable point) and the interval which is differentiable is (a,b), not [a,b] (3 votes) Avin 3 months agoPosted 3 months ago. Direct link to Avin's post “I'm a bit confused on the...” I'm a bit confused on the graph given by the initial explanation; the MVT says that the function must be continuous on the endpoints as well, though on the graph, the right sided limit does not exist for x=6, making it discontinuous over the interval [0,6]. Is that correct or am I confusing the concept? • (1 vote) WaGyu 3 months agoPosted 3 months ago. Direct link to WaGyu's post “Yeah, you have some confu...” Yeah, you have some confusion about continuity. (2 votes) mdabney 3 years agoPosted 3 years ago. Direct link to mdabney's post “The question in practice ...” The question in practice problem 1 specifically asks for a value for c in the closed interval [0,3], but marks one of the correct solutions as incorrect. I understand why it wouldn't want to accept 3 as a correct answer as it's the answer to one of the intermediate steps toward the solution it's looking for, but it is a correct answer to the question as it's currently posed. The question wording should be changed to ask for c in the open interval (0,3) to prevent that mistake. As Lori Feng mentions below, it's even described to be incorrect in the explanation because it assumes the question is asking for a value in the open interval rather than the closed interval listed in the question. • (1 vote) loumast17 3 years agoPosted 3 years ago. Direct link to loumast17's post “in the definition of the ...” in the definition of the mean value theorem given at the top it says that the interval [a,b] is for the function, then the interval (a,b) is for the derivative, which is where c would be. It is kinda crummy it's like that, but technically yeah, c would only be in the open interval (2 votes) Ray2017 6 months agoPosted 6 months ago. Direct link to Ray2017's post “Why do we exclude a and b...” Why do we exclude a and b when we want to find the possible values for c? • (1 vote) pufferfish56 4 days agoPosted 4 days ago. Direct link to pufferfish56's post “MVT only cares if the fun...” MVT only cares if the function is continuous at the endpoints, not any range outside of that (and thus doesn't care if the endpoints are differentiable), so if MVT holds true, there is at least one solution that is not one of the endpoints between a and b. That doesn't mean that the endpoints couldn't have a parallel secant line and tangent line, but it means it's not part of the range where it's guaranteed. (1 vote) zubair 5 years agoPosted 5 years ago. Direct link to zubair's post “if this man stop for some...” if this man stop for some minutes and continue his journey then? • (1 vote) jagetiyaareen5 4 years agoPosted 4 years ago. Direct link to jagetiyaareen5's post “Sir what about roll's and...” Sir what about roll's and lagrance's theorem • (1 vote) No_Solution 3 months agoPosted 3 months ago. Direct link to No_Solution's post “What would be an example ...” What would be an example where, if f is differentiable between a CLOSED interval of [a, b] instead of an OPEN interval (a,b), the MVT isn't valid? • (1 vote) kubleeka 3 months agoPosted 3 months ago. Direct link to kubleeka's post “(a, b) is a subset of [a,...” (a, b) is a subset of [a, b], so if f is differentiable on [a, b], then it is automatically differentiable on (a, b) as well. We use an open interval for that part of the statement because we generally like to use the weakest assumptions possible, so that we can apply the theorem in as many cases as possible. If I have a function that equals sin(x) for 0<x<π and 0 otherwise, then we can apply the mean value theorem on the interval [0, π], even though our function isn't differentiable at the endpoints. If we insisted on using the closed interval in the statement of the theorem, then we couldn't apply the theorem here, and for no good reason. (1 vote) Maha Rahman 6 years agoPosted 6 years ago. Direct link to Maha Rahman's post “If a=b, (the two endpoint...” If a=b, (the two endpoints have the same y value) can mean value theorem be applied? • (0 votes) kubleeka 6 years agoPosted 6 years ago. Direct link to kubleeka's post “Yes, it would imply that ...” Yes, it would imply that the function has a point in the interval where the derivative is 0. This special case is called Rolle's Theorem. (2 votes)Want to join the conversation?
A discontinuous point means that either you have no defined value at that point or it jumps to a different point. The graph you're talking doesn't show a jump at the point nor an undefined point (which is typically shown as an empty circle).
MVT doesn't care about differentiability at the endpoints as it states that it "must be differentiable over the open interval (a,b) and continuous over the closed interval [a,b]." Hope it helped.
and thank u sir
