The question

- mathmath333

The question

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- Here_to_Help15

The explanation

- 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

ooooo i remember this what are these called again???

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## More answers

- anonymous

x=0 =>y=8

- mathmath333

ok now

- Pawanyadav

K belongs to all real number as X is 0
Any real value of K don't affect the question.

- 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

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

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

only the set defined by system III is empty

- 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

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

at x=0 they return y=8
furthermore the first derivative of the first expression is zero when k=4

- mathmath333

is k=4 the answer

- Michele_Laino

yes!

- Michele_Laino

am I right?

- mathmath333

answer given is k=3

- Michele_Laino

sorry for my error then!

- mathmath333

is their any reverse method to use, in case the answer is known

- IrishBoy123

i think the answer is wrong
and i think it is "probably" 2

- 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

https://www.desmos.com/calculator/khqbigmxli

- Michele_Laino

thanks! :) @mathmath333

- BAdhi

so \(k\in \mathrm Z\) is another requirement right?

- 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

from the text of the problem I understand that:
\[k \in \mathbb{R}\]

- Michele_Laino

our function is not differentiable at x=0

- Michele_Laino

as we can see form the graph of @mathmath333

- 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

whereas the equation:
\[4 - k = - \left( {2 - k} \right)\]
has the right solution

- 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

@BAdhi
that is precisely how i saw it too

- BAdhi

In the link that @mathmath333 shared with us clearly shows that the answer is 2

- Michele_Laino

thanks! :) @BAdhi

- Michele_Laino

@IrishBoy123 I thought you were my friend, lol!

- ganeshie8

Interesting that for \(k\gt 4\), the min value occurs at \(x=\frac{2}{k}\)

- 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

except for k=2 and k=4 i think..

- ganeshie8

yeah for k=2 and k=4 we get a range for x, but above statement still holds i guess

- kanwal32

easy way draw a graph

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