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Numerade Educator



Problem 50 Hard Difficulty

Suppose that a function $ f $ is continuous on $ [0, 1] $ except at 0.25 and that $ f(0) = 1 $ and $ f(1) = 3 $. Let $ N = 2 $. Sketch two possible graphs of $ f $, one showing that $ f $ might not satisfy the conclusion of the Intermediate Value Theorem and one showing that $ f $ might still satisfy the conclusion of the Intermediate Value Theorem (even though it doesn't satisfy the hypothesis.)


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Video Transcript

and this is problem number fifty of the Stuart Calculus eighth edition, Section two point five. Suppose that a function F is continuous on zero two one, except at zero point two five, and that F zero equals one. An effort one equal stream let and equal to two sketch to possible grass F one showing that might not satisfy the conclusion of the Internet. Intermediate values here and one showing that right still satisfied the conclusion of the Intermediate Value theorem, even though it doesn't satisfy that hypothesis. So just a review. The intermediate Value three dirham states that on a certain interval a to B in this case Cyrano one for a function that is continuous between zero and one and the range between zero and one. Ah, that for any end between f zero on that one, there exists a scene in the interval zero one in this case that would ah be equal to that certain and comes of it show. In this case, we have a discontinues function at zero point two five, and we're going to survive this. It's not true. So after zero will be one half of Warner will be three and The only way that we're going to make this not true is by drawing a function. Where does this continuous add zero point two five? You were just going to put a hole but continuous, sir, in the remainder of the interval. So for this function, it does not satisfy the conclusion of the intermediate valiant Kerem because there is no value sea between zero and one or the function is equal to have two. And so we write that as there's the wife of scene that is equal to him. So this is an impossible on this graph we can share a grand for that might be true. Here again, on the right side, we have a zero point two five and her function Could I look a little different but still have a he dis continuity at there point of fire and it still had seven zero recall one for one, two, three. But this case, it's different because, and at an equals to two, which is this red line shown there does exist a see value in this case. Sara point seven five there exists. Is he value such that their function is equal to end or in this case, two so thes two crowds are examples of one that doesn't satisfy the condition in this case, notice that the whole that is continuity coincides with E um value. And therefore there is no choice of see where this is true But on this ah, graft to the right we kept the discontinuity, but we were also able to satisfy conclusion of the interview devalue film