Question

Explain in your own words how the Intermediate Value Theorem tells us that our intuitive idea that a continuous function on [a, b] should have no holes or jumps is correct. Illustrate your explanation with the help of the graph of a function f which is continuous on (-1, 1], and has a single critical point at x = 0. Give an example of a function which illustrates why f must be continuous at x = b in order to apply the Intermediate Value Theorem. Your graph must be continuous for all x # b.

Answer 1

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Intermediate value theorem ; - let, t be a continuous function defined on [a,b] and at st (flax, fio). Then, there exists cela,b) such that f(e)=s. Bje, f assume every value between fras fcb). LMz Suppose, f satisfies intermediate value theorem on the interras [a,b]. Then, I can't have holes as jump between flag f f(b). If possible lit, food has a jump at x=( Let it from = M (uts say mat) do te im = N (ut's say N= 11). According to intermediate value theorem, f assume every value between f(a) of f(b). consider the graph - ne- i soffb) flet)- ------ ----.--...--------- Forum 919 FC)=7- tra) n=l

The fonction is defined in the internal [a,b], we do not know f has any value outside of the internal [a,b] . ut's say it non no value outside tais] (ise does not exists.) ... Now, According to to intermediate value theorem, there must exists some point ce(a,b) such that f(1) = 9 (a point beth M&N) :M=4.. Nall value But from the graph we considered, there is not point, where f can take the value 9. so, intermediate value to fails here. If we want to make intermediate C But it is a contradiction, because we have assumed intermediate value holds. Hence our assumption that ar from=7, fr. fw=11 NC- is wrong. ije, it must be it tim = 4t fim. ise, fro) should be continuous. In another word, from intermediate value theorem, we have seen that a continuous function should have no holes.

on [-" I but nzo is Graphi- t is continuous an exition that point. AY fros = thi х! к -yx clearly n=0 is a critical point a singuler point . here, (-1)= -1 & fci) = 4. Sty of stand trave satisfied

a t must have to assume every ralues beth +(-1) f fed & ise, every value been 4 &1. Mows -1 <021. ise, there must enist some ct (1, 1) such that f(e) = 0 But, there is no point in C+1) for which (6)=0 holds. - Hence, fe sow the graph of f befn E'll should have no holes. In another word, If we consider intermadiate value theorem is true then, a continuous function should have no holes. - This complete the proof. - your Last Question is already quite Answered in above, we have taken here the point n=b=0 in [1,1]. X -