f cts on [a,b] and f(x)>0 for all x∈[a,b] implies that there exists an m>0 such that f(x)≥m for all x∈[a,b].

Kevin Charles

Kevin Charles

Answered question

2022-10-26

f cts on [ a , b ] and f ( x ) > 0 for all x [ a , b ] implies that there exists an m > 0 such that f ( x ) m for all x [ a , b ].
Proposition 1 seems to obviously be true. If f ( x ) is not equal to 0, it has to be positive or negative? Prop 2 seems to be false because what if the function approaches 0 as x approaches ?

Answer & Explanation

Plutbantonavv

Plutbantonavv

Beginner2022-10-27Added 15 answers

Extreme value theorem. (Note that x can’t approach : x [ a , b ])

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