Question

The question

Answer 1

The explanation

Answer 2

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

Answer 3

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

Answer 4

x=0 =>y=8

Answer 5

ok now

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

only the set defined by system III is empty

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

is k=4 the answer

Answer 15

yes!

Answer 16

am I right?

Answer 17

answer given is k=3

Answer 18

sorry for my error then!

Answer 19

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

Answer 20

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 ;-))

Answer 21

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

Answer 22

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

Answer 23

thanks! :) @mathmath333

Answer 24

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

Answer 25

@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

Answer 26

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

Answer 27

our function is not differentiable at x=0

Answer 28

as we can see form the graph of @mathmath333

Answer 29

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

Answer 30

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

Answer 31

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 \(y(0)\)

Answer 32

@BAdhi
that is precisely how i saw it too

Answer 33

In the link that @mathmath333 shared with us clearly shows that the answer is 2<k<4 also.

And @Michele_Laino wasnt wrong afterall ;)

Answer 34

thanks! :) @BAdhi

Answer 35

@IrishBoy123 I thought you were my friend, lol!

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

easy way draw a graph