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  1. #1
    Silver Lounger kweaver's Avatar
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    Find the minimum value

    What is the minimum value of f(x) = | -(x - h) + k | - q for each set of positive real numbers h, k, and q?

    (A) -q (B) -k (C) k (D) -k - q (E) k - q

    While there's a 20% chance of guessing correctly, it's necessary to "explain" your answer.

    This was a question on a recent standardized test for high school students.

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    Super Moderator BATcher's Avatar
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    Possibly the high-school students had the benefit of mathematical typesetting - unfortunately I find the problem unintelligible as presented...
    BATcher

    Existence is useful.

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    Silver Lounger kweaver's Avatar
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    Here's a graphic of f(x) = | -(x - h) + k | - q

    Clip0001.jpg

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    Super Moderator BATcher's Avatar
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    The problem is therefore that I don't have a clue what the vertical bars represent!

    And the other problem is that I don't know what
    (A) -q (B) -k (C) k (D) -k - q (E) k - q
    means...

    Perhaps I shouldn't have started this?
    BATcher

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    WS Lounge VIP access-mdb's Avatar
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    I assume that this is the modulus of the result - see Wikipedia. But like you Batcher I haven't a clue either
    What do you mean nothing is impossible? I've been doing nothing for years.

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    WS Lounge VIP Browni's Avatar
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    Quote Originally Posted by BATcher View Post
    And the other problem is that I don't know what
    (A) -q (B) -k (C) k (D) -k - q (E) k - q
    means...
    Multiple choice answers

    A. -q
    B. -k
    C. k
    D. -k-q
    E. k-q

    The clue was in the 1 in 5 chance of getting it right!
    Last edited by Browni; 2017-02-18 at 16:54.

  7. #7
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    I take the vertical bars to mean "absolute value" (i.e., if the expression inside the vertical bars evaluates to a negative value, multiply it by -1). I'll defer posting my answer.

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    Silver Lounger kweaver's Avatar
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    The vertical bars mean ABSOLUTE VALUE. That's common mathematical notation.

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    To minimize f(x) = | -(x - h) + k | - q, try to set x such that taking the absolute value yields the lowest possible value, i.e., zero:

    -(x - h) + k = 0,
    -(x - h) = -k,
    (x - h) = k,
    x - h = +/-sqrt(k), # "+/-" represents "plus or minus"
    x = +/-sqrt(k) + h,

    which is "real" because k is a positive real number.

    So, the answer is (A) -q.

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    Silver Lounger kweaver's Avatar
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    how do you know for sure that the absolute value of the quadratic is = 0?

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    I'm not sure I understand the question, but I believe I showed that the quadratic is zero if and only if x = +/-sqrt(k) + h. Therefore, setting x = +/-sqrt(k) + h will result in the quadratic's being zero.

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    I'm not sure why you need to bother with what x might be: the absolute value part must be +ve, its minimum value must therefore be zero (for whatever x, h or k), and if we take q away from zero the minimum must be -q.

    What age might these students be?
    Last edited by mngerhold; 2017-02-23 at 12:22.

  13. #13
    Silver Lounger kweaver's Avatar
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    Mngerhold:

    Standardized test(s) given to (generally) 11th grade students.

    You are correct about your answer and approach. It's actually fairly straight-forward.

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    The solution is E k-q. If x=h then (x-h)=0>>0^2=0 >>| -0+k | -q = [abs val of -k ] -q = k-q which is choice E.

    Would you believe this is Algebra 1 ?

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    I'm not seeing how setting x to h satisfies the problem of minimizing f(x). As I wrote earlier, that problem seems to require getting the entire expression within the absolute value symbols to equal zero.

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