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mathmath333
 one year ago
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mathmath333
 one year ago
The question

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Here_to_Help15
 one year ago
Best ResponseYou've already chosen the best response.0The explanation

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align}& y=3x+6+x+kx2\hspace{.33em}\\~\\ &y_{\text{min}}\ \ \text{is at} \ \ x=0\ \ \text{and }\ k>0\hspace{.33em}\\~\\ &\text{Find } \ k\hspace{.33em}\\~\\ \end{align}}\)

Here_to_Help15
 one year ago
Best ResponseYou've already chosen the best response.0ooooo i remember this what are these called again???

Pawanyadav
 one year ago
Best ResponseYou've already chosen the best response.0K belongs to all real number as X is 0 Any real value of K don't affect the question.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2only the set defined by system III is empty

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2now, 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
 one year ago
Best ResponseYou've already chosen the best response.2at x=0 they return y=8 furthermore the first derivative of the first expression is zero when k=4

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2answer given is k=3

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2sorry for my error then!

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2is their any reverse method to use, in case the answer is known

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i 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 ;))

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2thanks! :) @mathmath333

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1so \(k\in \mathrm Z\) is another requirement right?

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.2from the text of the problem I understand that: \[k \in \mathbb{R}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2our function is not differentiable at x=0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2as we can see form the graph of @mathmath333

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for example the first derivative of the finction: \[y = 4x + 6  kx + 2,\quad x \in I\] is 4k whereas the first derivative of the function: \[y = 2x + 6  kx + 2,\quad x \in II\] is 2k and the equation 4k=2k has no solutions

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2whereas the equation: \[4  k =  \left( {2  k} \right)\] has the right solution

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1I think it should be like this. when \(2 < x< 0\) \(y =(2k)x+8\) to get the minimum at y(0) the slop of this function should be less than zero (since x<0) thus, \(y' = (2k) < 0 \implies k > 2\) when \(x>0\) \(y= (4k)+8\) to get minimum at y(0) the slop of this function should be larger than zero (x>0) thus, \(y' = (4k) >0 \implies k<4\) so for any value \(2<k<4\) the function will give minimum value at \(y(0)\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@BAdhi that is precisely how i saw it too

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1In the link that @mathmath333 shared with us clearly shows that the answer is 2<k<4 also. And @Michele_Laino wasnt wrong afterall ;)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2thanks! :) @BAdhi

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2@IrishBoy123 I thought you were my friend, lol!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Interesting that for \(k\gt 4\), the min value occurs at \(x=\frac{2}{k}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1geometrically, i think it helps to see that min value always occurs at one of the turning points : \[\{2,~0,~\frac{2}{k}\}\]

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1except for k=2 and k=4 i think..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yeah for k=2 and k=4 we get a range for x, but above statement still holds i guess

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0easy way draw a graph
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