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Prove f has maximum value in R

  1. Apr 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Let f: ℝ→ℝ be continuous function such that f(x) approaches 0 when x---> -∞ and f(0)=2. In addition, f is decreasing when x≥3. Prove f has a maximum (greatest value) in ℝ.

    2. Relevant equations
    Theorem: If function f is continuous on a closed interval [a, b], then f has maximum and minimum in [a, b].

    3. The attempt at a solution

    Because f(x) approaches 0 when x ---> -∞ then there exists a<0 such that a<0 ⇒ |f(x)-0|=|f(x)|<2. So a<0 ⇒ f(x)<2. Because f is continuous in ℝ, it's also continuous on a closed interval [a,3]⊂ℝ and there's the theorem in our lecture notes that says now ∃x0∈[a,3] such that f(x)≤f(x0) ∀x∈[a,3]. Since a is negative, 0∈[a,3] and therefore f(x0)≥f(0)=2. Because f is decreasing when x≥3, f(x)≤f(x0), when x>x0≥3.

    The idea is to divide ℝ into intervals so that we get closed interval to apply the theorem.

    In a summary:
    ##a<0 ⇒ f(x)<2≤f(x_0)##
    ##x∈[a,3] ⇒f(x)≤f(x_0)##
    ##x≥3 ⇒f(x)≤f(x_0)##

    ∴ f has greatest value/maxima in ℝ.

    Is this proof correct and rigorious enough? I think I have made a mistake in the end when it comes to the fact that f is decreasing when x≥3.
     
    Last edited: Apr 12, 2016
  2. jcsd
  3. Apr 12, 2016 #2

    LCKurtz

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    I think you have the right idea. I would change your last inequality to$$
    x\ge 3 \implies f(x) \le f(3) \le f(x_0)$$to make it explicit where you use the fact that ##f## is decreasing on ##[3,\infty)##.
     
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