[SOLVED] Max and point 2, absolute value

[Petrus]

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]

\(\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}\)

 

[Petrus]

\(\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}\)

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]

Do you mean the domain ? Because the domain represents the values for x but the range represents the values for y

 

[Petrus]

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 ? Sorry I did not know it's called domain in english

 

[ZaidAlyafey]

Ok , then you are correct , derive each function separately .

 

[Petrus]

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]

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]

Hello,
Thanks ZaidAlyafey and Jameson for the help! I did correct answer!
after input all those point into orginal function and look for highest value ( not wealth ) 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?

 

[Jameson]

Cool So you're saying you've confirmed that you have found the correct answer?

 

[Petrus]

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]

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