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Let $f(x)$ be a continuous function in $[1,3]$ defined for all $x$ belonging to $ R $. If $f(x)$ take rational values for all $x$ belonging to R and $f(2)=198$ then $f(2^2)$

Aakash Kumar
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  • Can you share what you've tried and what you're having trouble with? This is essentially a very standard exercise in applying some theorem about intermediate values. –  Jun 11 '16 at 06:40
  • I really don't get what to do. – Aakash Kumar Jun 11 '16 at 06:43
  • i dont think there is one solution , question is incomplete for ex: take $f(x)=198$ , for all $x$ or $f(x)= 196+x$ and so on. i would suggest typing the whole question from whichever source you are typing, and post what ideas you had when you saw the question – avz2611 Jun 11 '16 at 06:44
  • Ok, then. My crystal ball says that you got this in a course immediately after talking about a theorem that says the words "intermediate value." If not, then tell us where you found this problem. –  Jun 11 '16 at 06:44
  • @avz2611 Does $196 + x$ have only rational outputs? –  Jun 11 '16 at 06:45
  • @T.Bongers it never said anything about rational outputs , it said function is capable of taking rational inputs – avz2611 Jun 11 '16 at 06:46
  • @avz2611 The question says that $f$ takes rational values for all inputs. –  Jun 11 '16 at 06:46
  • ah i misunderstood the question you are right – avz2611 Jun 11 '16 at 06:47
  • I found it in my book – Aakash Kumar Jun 11 '16 at 06:49
  • The solution is continuous in [1,3] – Aakash Kumar Jun 11 '16 at 06:53
  • Sorry function is continuous in [1,3] – Aakash Kumar Jun 11 '16 at 06:54

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I guess f(4)=198 also, bechause if it were any other value, then by the intermidiate value theorem, it would take all values between 198 and said other value. I don't, however, understand why f is only given to be continious in [1;3] when we are asked about f(4)

Probability Boi
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