## mathmath333 one year ago The question

1. Here_to_Help15

The explanation

2. 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}}

3. Here_to_Help15

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

4. anonymous

x=0 =>y=8

5. mathmath333

ok now

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

7. 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

8. 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

9. Michele_Laino

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

10. Michele_Laino

only the set defined by system III is empty

11. Michele_Laino

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

12. Michele_Laino

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

13. Michele_Laino

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

14. mathmath333

15. Michele_Laino

yes!

16. Michele_Laino

am I right?

17. mathmath333

18. Michele_Laino

sorry for my error then!

19. mathmath333

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

20. IrishBoy123

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

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
23. Michele_Laino

thanks! :) @mathmath333

so $$k\in \mathrm Z$$ is another requirement right?

@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

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

27. Michele_Laino

our function is not differentiable at x=0

28. Michele_Laino

as we can see form the graph of @mathmath333

29. 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

30. Michele_Laino

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

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

@BAdhi that is precisely how i saw it too

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

35. Michele_Laino

@IrishBoy123 I thought you were my friend, lol!

36. ganeshie8

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

37. ganeshie8

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

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

39. ganeshie8

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

40. kanwal32

easy way draw a graph