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Inequality Proof

  1. Aug 12, 2010 #1
    1. The problem statement, all variables and given/known data

    If f(x) = (x-1)^2 and g(x) = x+1, then g is greater than or equal to f on the set S = {real numbers x : x is between 0 and 3}.

    2. Relevant equations

    g is greater than or equal to f on the set S of real numbers iff for all s in S, g(s) is greater than f(s).

    3. The attempt at a solution

    Since we know x is an element of S, we know that x is between 0 and 3. That is, (x)(x-3) is less than or equal to 0. And here is where I get stuck.

    I have tons of scratch paper that doesn't really show anything, and my TA gave what he calls a "proof" of this, but he assumed f(x) is less than or equal to g(x), but doesn't realize that assuming what you are trying to prove is not a way to prove anything. I just don't know how to start this.
    Last edited: Aug 12, 2010
  2. jcsd
  3. Aug 12, 2010 #2


    Staff: Mentor

    Hopefully, you have sketched a graph of both functions. If so, you should see that the two curves intersect at (0, 1) and (3, 4).

    Look at the expression x + 1 - (x - 1)2. On any interval where this expression is positive, g(x) > f(x). On any interval where the expression is negative, g(x) < f(x). Note that I am not a priori assuming either function is larger than the other.
  4. Aug 12, 2010 #3
    For some reason I never thought to subtract f(x) from g(x). It is proved! Thank you!
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