Answer to Question #184134 in Real Analysis for Maria

Question #184134

Find the infimum and supremum in each of the following sets of real numbers: S = {x| − x2 + 6x − 3 > 0


(ii) Let a be the supremum of a set of real numbers and let ε > 0 be any real number.Show that there is at least one x ∈ S such that a−ε<x≤a where S is the set with the given supremum.


1
Expert's answer
2021-04-25T17:01:53-0400

(i) Given set is-

"S = x: \u2212 x^2 + 6x \u2212 3 > 0"


"\\Rightarrow x^2-6x+3<0\\\\\n\\Rightarrow (x-(3+\\sqrt{6})(x-(3-\\sqrt{6})<0"


This will only possible when-

"(x-(3+\\sqrt{6}))<0 \\text{ and }(x-(3-\\sqrt{6})>0"


"\\implies 3-\\sqrt{6}<x<3+\\sqrt{6}"


Hence inf{S}"=3-\\sqrt{6}"

sup{S}"=3+\\sqrt{6}"


(ii) As sup{S}=a, where S is set of real numbers.

As "\\epsilon >0" , Implies that There always exist a supremum between the "a-\\epsilon" and "a" , As a contain the supremum before the "\\epsilon" came into picture.

Hence there is at least one x ∈ S such that "a\u2212\\epsilon<x\u2264a" where S is the set with the given supremum.


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Comments

Assignment Expert
10.05.21, 10:34

Dear Donna, please use the panel for submitting a new question. The statement of a new question is incomplete. What should be done there?

Donna
03.05.21, 14:57

Let f (x) be defined as follows f(x) ={x2+8x+15 /x+3 ifx

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