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mathmath333

  • one year ago

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  1. Here_to_Help15
    • one year ago
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    The explanation

  2. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& y=|3x+6|+|x|+|kx-2|\hspace{.33em}\\~\\ &y_{\text{min}}\ \ \text{is at} \ \ x=0\ \ \text{and }\ k>0\hspace{.33em}\\~\\ &\text{Find } \ k\hspace{.33em}\\~\\ \end{align}}\)

  3. Here_to_Help15
    • one year ago
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    ooooo i remember this what are these called again???

  4. anonymous
    • one year ago
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    x=0 =>y=8

  5. mathmath333
    • one year ago
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    ok now

  6. Pawanyadav
    • one year ago
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    K belongs to all real number as X is 0 Any real value of K don't affect the question.

  7. IrishBoy123
    • one year ago
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    you can establish that k>2 for starters as you need a negative slope just to left of y axis for x = 0 to be min you can also establish that k > 4 "just to" right hand to have +ve slope, but it gets more messy as you move further away in he +ve x direction past the intercept y = 0, x = 2/k

  8. mathmath333
    • one year ago
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    The assumption that any real value \(k\) works is wrong for example at \(k=4\) \(y_{\text{min}}\) is same for the inequality \(0\leq x\leq 0.5\) https://www.desmos.com/calculator/hrm0jyq1zg

  9. Michele_Laino
    • one year ago
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    I consider these four cases: \[\begin{gathered} \left\{ \begin{gathered} 3x + 6 \geqslant 0 \hfill \\ x \geqslant 0 \hfill \\ \end{gathered} \right.I,\quad \left\{ \begin{gathered} 3x + 6 \geqslant 0 \hfill \\ x < 0 \hfill \\ \end{gathered} \right.II \hfill \\ \left\{ \begin{gathered} 3x + 6 < 0 \hfill \\ x \geqslant 0 \hfill \\ \end{gathered} \right.III,\quad \left\{ \begin{gathered} 3x + 6 < 0 \hfill \\ x < 0 \hfill \\ \end{gathered} \right.IV \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    only the set defined by system III is empty

  11. Michele_Laino
    • one year ago
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    for all x belonging to sets I and II, we have: \[\begin{gathered} y = 4x + 6 + \left| {kx - 2} \right|,\quad x \in I \hfill \\ y = 2x + 6 + \left| {kx - 2} \right|,\quad x \in II \hfill \\ \end{gathered} \]

  12. Michele_Laino
    • one year ago
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    now, for both expressions we consider \[kx - 2 < 0\] so, we can write: \[\begin{gathered} y = 4x + 6 - kx + 2,\quad x \in I \hfill \\ y = 2x + 6 - kx + 2,\quad x \in II \hfill \\ \end{gathered} \]

  13. Michele_Laino
    • one year ago
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    at x=0 they return y=8 furthermore the first derivative of the first expression is zero when k=4

  14. mathmath333
    • one year ago
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    is k=4 the answer

  15. Michele_Laino
    • one year ago
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    yes!

  16. Michele_Laino
    • one year ago
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    am I right?

  17. mathmath333
    • one year ago
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    answer given is k=3

  18. Michele_Laino
    • one year ago
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    sorry for my error then!

  19. mathmath333
    • one year ago
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    is their any reverse method to use, in case the answer is known

  20. IrishBoy123
    • one year ago
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    i think the answer is wrong and i think it is "probably" 2<k<4 i have run a short analytic on it which thus far comfirms this this is also borne out by just looking at the gradients on either side of the y axis which yields these 2 consitions. i would also be surprised to find @Michele_Laino in error ;-))

  21. Michele_Laino
    • one year ago
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    I think that my expressions \[\begin{gathered} y = 4x + 6 - kx + 2,\quad x \in I \hfill \\ y = 2x + 6 - kx + 2,\quad x \in II \hfill \\ \end{gathered} \] are correct, nevertheless the answer k=4, is wrong @IrishBoy123

  22. mathmath333
    • one year ago
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    https://www.desmos.com/calculator/khqbigmxli

  23. Michele_Laino
    • one year ago
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    thanks! :) @mathmath333

  24. BAdhi
    • one year ago
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    so \(k\in \mathrm Z\) is another requirement right?

  25. BAdhi
    • one year ago
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    @Michele_Laino you have stopped in the middle of the proof. From first expression you get \(k < 4\). From second expression you get \(k>2\) If \(k\in \mathrm Z\) only possible answer is k=3

  26. Michele_Laino
    • one year ago
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    from the text of the problem I understand that: \[k \in \mathbb{R}\]

  27. Michele_Laino
    • one year ago
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    our function is not differentiable at x=0

  28. Michele_Laino
    • one year ago
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    as we can see form the graph of @mathmath333

  29. Michele_Laino
    • one year ago
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    for example the first derivative of the finction: \[y = 4x + 6 - kx + 2,\quad x \in I\] is 4-k whereas the first derivative of the function: \[y = 2x + 6 - kx + 2,\quad x \in II\] is 2-k and the equation 4-k=2-k has no solutions

  30. Michele_Laino
    • one year ago
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    whereas the equation: \[4 - k = - \left( {2 - k} \right)\] has the right solution

  31. BAdhi
    • one year ago
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    I think it should be like this. when \(-2 < x< 0\) \(y =(2-k)x+8\) to get the minimum at y(0) the slop of this function should be less than zero (since x<0) thus, \(y' = (2-k) < 0 \implies k > 2\) when \(x>0\) \(y= (4-k)+8\) to get minimum at y(0) the slop of this function should be larger than zero (x>0) thus, \(y' = (4-k) >0 \implies k<4\) so for any value \(2<k<4\) the function will give minimum value at \(y(0)\)

  32. IrishBoy123
    • one year ago
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    @BAdhi that is precisely how i saw it too

  33. BAdhi
    • one year ago
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    In the link that @mathmath333 shared with us clearly shows that the answer is 2<k<4 also. And @Michele_Laino wasnt wrong afterall ;)

  34. Michele_Laino
    • one year ago
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    thanks! :) @BAdhi

  35. Michele_Laino
    • one year ago
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    @IrishBoy123 I thought you were my friend, lol!

  36. ganeshie8
    • one year ago
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    Interesting that for \(k\gt 4\), the min value occurs at \(x=\frac{2}{k}\)

  37. ganeshie8
    • one year ago
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    geometrically, i think it helps to see that min value always occurs at one of the turning points : \[\{-2,~0,~\frac{2}{k}\}\]

  38. BAdhi
    • one year ago
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    except for k=2 and k=4 i think..

  39. ganeshie8
    • one year ago
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    yeah for k=2 and k=4 we get a range for x, but above statement still holds i guess

  40. kanwal32
    • one year ago
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    easy way draw a graph

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