mathmath333
  • mathmath333
The question
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Here_to_Help15
  • Here_to_Help15
The explanation
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align}& y=|3x+6|+|x|+|kx-2|\hspace{.33em}\\~\\ &y_{\text{min}}\ \ \text{is at} \ \ x=0\ \ \text{and }\ k>0\hspace{.33em}\\~\\ &\text{Find } \ k\hspace{.33em}\\~\\ \end{align}}\)
Here_to_Help15
  • Here_to_Help15
ooooo i remember this what are these called again???

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More answers

anonymous
  • anonymous
x=0 =>y=8
mathmath333
  • mathmath333
ok now
Pawanyadav
  • Pawanyadav
K belongs to all real number as X is 0 Any real value of K don't affect the question.
IrishBoy123
  • IrishBoy123
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
mathmath333
  • mathmath333
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
Michele_Laino
  • Michele_Laino
I consider these four cases: \[\begin{gathered} \left\{ \begin{gathered} 3x + 6 \geqslant 0 \hfill \\ x \geqslant 0 \hfill \\ \end{gathered} \right.I,\quad \left\{ \begin{gathered} 3x + 6 \geqslant 0 \hfill \\ x < 0 \hfill \\ \end{gathered} \right.II \hfill \\ \left\{ \begin{gathered} 3x + 6 < 0 \hfill \\ x \geqslant 0 \hfill \\ \end{gathered} \right.III,\quad \left\{ \begin{gathered} 3x + 6 < 0 \hfill \\ x < 0 \hfill \\ \end{gathered} \right.IV \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
only the set defined by system III is empty
Michele_Laino
  • Michele_Laino
for all x belonging to sets I and II, we have: \[\begin{gathered} y = 4x + 6 + \left| {kx - 2} \right|,\quad x \in I \hfill \\ y = 2x + 6 + \left| {kx - 2} \right|,\quad x \in II \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now, for both expressions we consider \[kx - 2 < 0\] so, we can write: \[\begin{gathered} y = 4x + 6 - kx + 2,\quad x \in I \hfill \\ y = 2x + 6 - kx + 2,\quad x \in II \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
at x=0 they return y=8 furthermore the first derivative of the first expression is zero when k=4
mathmath333
  • mathmath333
is k=4 the answer
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
am I right?
mathmath333
  • mathmath333
answer given is k=3
Michele_Laino
  • Michele_Laino
sorry for my error then!
mathmath333
  • mathmath333
is their any reverse method to use, in case the answer is known
IrishBoy123
  • IrishBoy123
i think the answer is wrong and i think it is "probably" 2
Michele_Laino
  • Michele_Laino
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
mathmath333
  • mathmath333
https://www.desmos.com/calculator/khqbigmxli
Michele_Laino
  • Michele_Laino
thanks! :) @mathmath333
BAdhi
  • BAdhi
so \(k\in \mathrm Z\) is another requirement right?
BAdhi
  • BAdhi
@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
Michele_Laino
  • Michele_Laino
from the text of the problem I understand that: \[k \in \mathbb{R}\]
Michele_Laino
  • Michele_Laino
our function is not differentiable at x=0
Michele_Laino
  • Michele_Laino
as we can see form the graph of @mathmath333
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
whereas the equation: \[4 - k = - \left( {2 - k} \right)\] has the right solution
BAdhi
  • BAdhi
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
IrishBoy123
  • IrishBoy123
@BAdhi that is precisely how i saw it too
BAdhi
  • BAdhi
In the link that @mathmath333 shared with us clearly shows that the answer is 2
Michele_Laino
  • Michele_Laino
thanks! :) @BAdhi
Michele_Laino
  • Michele_Laino
@IrishBoy123 I thought you were my friend, lol!
ganeshie8
  • ganeshie8
Interesting that for \(k\gt 4\), the min value occurs at \(x=\frac{2}{k}\)
ganeshie8
  • ganeshie8
geometrically, i think it helps to see that min value always occurs at one of the turning points : \[\{-2,~0,~\frac{2}{k}\}\]
BAdhi
  • BAdhi
except for k=2 and k=4 i think..
ganeshie8
  • ganeshie8
yeah for k=2 and k=4 we get a range for x, but above statement still holds i guess
kanwal32
  • kanwal32
easy way draw a graph

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