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[SOLVED] Max and point 2, absolute value

Petrus

Well-known member
Feb 21, 2013
739
Calculate max and min point to function \(\displaystyle \frac{x^3}{2}-|1-4x|\) in the range \(\displaystyle \left(0,2 \right)\)
I got one question, shall I ignore when it's \(\displaystyle \frac{x^3}{2}-(-1+4x)\) cause then \(\displaystyle x<0\) and that dont fit in my range? Do I got correct?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}\)
 

Petrus

Well-known member
Feb 21, 2013
739
\(\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}\)
So I basicly got wrong? (I have not done this kind of problem with aboslute value)
If I understand correct so I will have
\(\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}\) that means i got new range on that? \(\displaystyle \left(\frac{1}{4},2 \right)\)
\(\displaystyle |1-4x|=-(-1+4) \,\,\, \,\,\,\forall \,\,\ x<\frac{1}{4} \) that means my new range is \(\displaystyle \left(0,\frac{1}{4} \right)\) I am right? If so shall I derivate both and find crit point?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Do you mean the domain ? Because the domain represents the values for x but the range represents the values for y
 

Petrus

Well-known member
Feb 21, 2013
739
Do you mean the domain ? Because the domain represents the values for x but the range represents the values for y
Yeah I mean domain, Thats what the problem is asking for :p? Sorry I did not know it's called domain in english
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Ok , then you are correct , derive each function separately .
 

Petrus

Well-known member
Feb 21, 2013
739
let's make them to case 1 and case 2
Case 1:
\(\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}\) \(\displaystyle \left(\frac{1}{4},2 \right)\)
Our function become \(\displaystyle \frac{x^3}{2}-1+4x\)
If we derivate the function we get \(\displaystyle \frac{3x^2}{2}+4\)
Now we want to find critical point \(\displaystyle \frac{3x^2}{2}+4=0\)
So I factor out \(\displaystyle \frac{1}{2}\) and get \(\displaystyle \frac{1}{2}(3x^2+8)=0\) and our critical point (complex) are \(\displaystyle x_1=\sqrt{\frac{-8}{3}}\) and \(\displaystyle x_2=-\sqrt{\frac{-8}{3}}\) I guess we shall ignore case 1 cause we got complex roots.
Case 2:
\(\displaystyle |1-4x|=-(-1+4) \,\,\, \,\,\,\forall \,\,\ x<\frac{1}{4} \) \(\displaystyle \left(0,\frac{1}{4} \right)\)
Our function become \(\displaystyle \frac{x^3}{2}+1-4x\)
If we derivate the function we get \(\displaystyle \frac{3x^2}{2}-4\)
Now we want to find critical point \(\displaystyle \frac{3x^2}{2}-4=0\)
So I factor out \(\displaystyle \frac{1}{2}\) and get \(\displaystyle \frac{1}{2}(3x^2-8)=0\) and our critical point are: \(\displaystyle x_1:\sqrt{\frac{8}{3}}\) and \(\displaystyle x_2:-\sqrt{\frac{8}{3}}\) but our \(\displaystyle x_2\) does not fit our domain so we shall ignore it. I am correct so far? What shall I do next?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Hi Petrus. It looks pretty good to me, but also hope someone else checks for any errors. What you should do now is plug in a bunch of stuff into the original f(x). When you have a restricted domain and you're trying to find minimums and maximums you should also test out:

1) Critical points
2) End points of the domain

So in this case you found just one critical point, \(\displaystyle x=\sqrt{\frac{8}{3}}\). Find the following values and then determine where you have a maximum and minimum. \(\displaystyle f \left( \sqrt{\frac{8}{3}} \right), \hspace{1 mm} f(0), \hspace{1 mm} f \left( \frac{1}{4} \right), \hspace{1 mm} f(2)\)
 

Petrus

Well-known member
Feb 21, 2013
739
Hello,
Thanks ZaidAlyafey and Jameson for the help! I did correct answer!:)(Bow)(Dance)
after input all those point into orginal function and look for highest value ( not wealth :p) and lowest value, I get the answer
max: \(\displaystyle \frac{1}{128}\)
min: \(\displaystyle 1-\frac{16\sqrt{\frac{2}{3}}}{3}\)
edit: what does \(\displaystyle \,\,\, \,\,\,\forall \,\,\,\) means in words?
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Cool :) So you're saying you've confirmed that you have found the correct answer?
 

Petrus

Well-known member
Feb 21, 2013
739
Cool :) So you're saying you've confirmed that you have found the correct answer?
Hello Jameson
Indeed I did thanks to you and ZaidAlyafey ( problem we get is online problem that means we put in our answer and it give us 'correct' or 'wrong' and I got 'correct') I also edit my early post if anyone could tell me what \(\displaystyle \,\,\, \,\,\,\forall \,\,\,\) that means in words :)
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Great! I'm going to mark this thread [SOLVED] then.

$\forall$ means "for all" or "for every". Earlier you wrote this:

\(\displaystyle \forall x>\frac{1}{4} \left(\frac{1}{4},2 \right)\)
which is a bit confusing. I think I know what you meant though and in my opinion it's better and more conventional to write it like this:

i) \(\displaystyle \forall x \in \left[ \frac{1}{4},2 \right] \) or

ii) \(\displaystyle x \mid \frac{1}{4}\le x \le 2\)