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Math Help - increase decrease constant

  1. #1
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    increase decrease constant

    Using my calculator I can find the answer but am having problems algebraically.

    -abs(x+4)-abs(x+1)

    If i take the first derivative and then set that equal to zero I get the numbers that I want.

    So I get

    -(x+4)/abs(x+4) - (x+1)/abs(x+1) How should I solve this. I am having a problem getting a common denominator. I do not know what to do with the absolute value signs.
    Thanks for the help.

    I know that x=-1.06383 and -3.829787 but I want to do this by hand and not the calculator.
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  2. #2
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    This is a min-max problem, right?

    The simplest way to do this problem is to graph it. Note that your first derivative does not exist at the points x = -4 and x = -1. However note that it does exist for any value in between x = -4 and x = -1. So there are not two solutions to this problem, but a whole interval on the x axis. (I'm not sure of where you got the "correct" answers, but they are not correct.)

    -Dan
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  3. #3
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    I used the calculator and trached the graph. So I have relative max at -4 and -1. constant between those values. increase until -4 and decrease after -1. When I trace I just used the approx values on the TI 84 that it gives. When I plugged them into the derivative it gave me zero. To find the turning point the first derivative equals zero, correct? When I used those values it gives me zero.
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  4. #4
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    Quote Originally Posted by IDontunderstand View Post
    Using my calculator I can find the answer but am having problems algebraically.

    -abs(x+4)-abs(x+1)

    If i take the first derivative and then set that equal to zero I get the numbers that I want.

    So I get

    -(x+4)/abs(x+4) - (x+1)/abs(x+1) How should I solve this. I am having a problem getting a common denominator. I do not know what to do with the absolute value signs.
    Thanks for the help.
    That's an awkward way to right the derivative!
    if x> -1, then both x+ 1 and x+ 4 are positive so both those fractions are simply "1". The derivative is -2. If -4< x< -1, then x+ 1 is negative but x+ 4 is still positive. Now the first fraction is 1 and the second one -1. Now their sum is 0. Finally, if x< -4, both fractions are -1 and so the derivative is 2:
    f(x)= \left\{\begin{array}{cc}2 & x< -4 \\ 0 & -4< x< -1 \\ -2 & -1< x\end{array}\right

    I know that x=-1.06383 and -3.829787 but I want to do this by hand and not the calculator.
    I am curious as to HOW you did that on a calculator. Going back to your orignal formula, -|x+4|-|x+1|, if x< -4, both quantities inside the absolute value signs are negative so this is -(-(x+4))-(-(x+1)= x+ 4+ x+ 1=2x+ 5 For -4< x< -1, x+ 4 posiive while x+ 1 is still negative so -(x+4)-(-(x+1))= -x- 4+ x+ 1= -3. Finally, for x> -1, both x+ 4 and x+ 1 are positive so -(x+4)- (x+1)= -2x- 5.

    It looks to me like you have become too focused on applying formulas and forgot to think about what absolute value means.

    The graph consists of three straight lines, a line with slope 2 going up to (-4, -3), a horizontal line from (-4, -3) to (-1, -3) then a line with slope -2 down from (-1, -3). The maximum value is -3 which occurs for all x between -4 and -1.
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