1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Problem on quadratic functions

  1. Jul 26, 2013 #1
    1. The problem statement, all variables and given/known data

    a) Let f(x) = ax^2+ bx + c; ∀x ∈ ℝ; where a(≠ 0), b and c ∈ ℝ are constants,
    (i) Prove that f(k) =(64+(b^2−4ac)^2)/(64a) ; where k = −(b/2a)+(1/a)+(b^2−4ac)/(8a).
    (ii) Hence, without using graphs and without using your knowledge on quadratic
    functions and equations, prove that (∀x ∈ ℝ f(x) > 0) [itex]\Rightarrow[/itex] (a > 0 and b^2 − 4ac< 0).

    2. Relevant equations



    3. The attempt at a solution
    i have already proven the first part but i m stuck in the 2nd part that says prove from "hence" . otherwise i can show that there exist M in R such that for all x in R ,M<=f(x) and M=inf(f(x)), and there exist x' in R such that f(x')=M, so i can take an argument that x' to be in real and if this happens (a>0 and b^2-4ac<0) ,M>0, so f(x)<0, but on that approach i m not using any above proven things to get my second answer(hence). if any one can help would be great. sorry for my english.
     
  2. jcsd
  3. Jul 26, 2013 #2
    You want to prove that the statement ∀x ∈ ℝ f(x) > 0) [itex]\Rightarrow[/itex] (a > 0 and b^2 − 4ac< 0) is a true statement. In this paticular case, if I were you I would start by thinking about when/under what conditions this statement isn't true. Do you know when an implication (p implies q) isn't true?

    Then I would try to find a contradiction. Are you familiar with proof by contradiction? If not, I would be happy to explain.

    This probably isn't the only way to do this, but it's the first solution I thought of and it seems well suited to this problem (I checked that the method works here).

    Dods
     
  4. Jul 26, 2013 #3
    thanks for the reply , p[itex]\Rightarrow[/itex]q is false only when the p is true and q is false. the problem I'm having is i cant find a way to say it from using f(k). it just a constant right ?. i cant use it to talk show that for all x f(x) is greater than 0. you mean that I should show the p (for all x f(x)>0 ) implies the negation of (a>0 & b^2-4ac<0) proposition is always false right ?
     
  5. Jul 26, 2013 #4
    I'm sorry, I didn't get the emphasis on "hence"...I just suggested a less convoluted way of proving it (if you take (∀x ∈ ℝ f(x) > 0) to be true and (a > 0 and b^2 − 4ac< 0) to be false you can easily arrive at a contradiction). I'll look at it again, although off the top of my head I don't see a connection to the part you've already proved.

    Can you clarify the context here? What sort of level is this book/course? Have you recently covered a method that could be relevant?

    I'll try to help as much as I can, however there are plenty of people waaay smarter than me on these forums who might be better able to help, or see something I'm missing.
     
  6. Jul 26, 2013 #5
    its on Mathematics of Bsc Engineering , level 1. its came up with the real analysis topic of functions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A Problem on quadratic functions
Loading...