- Thread starter
- #1

#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

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?

- Thread starter Petrus
- Start date

- Thread starter
- #1

- Feb 21, 2013

- 739

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?

- Jan 17, 2013

- 1,667

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

- Thread starter
- #3

- Feb 21, 2013

- 739

So I basicly got wrong? (I have not done this kind of problem with aboslute value)\(\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}\)

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?

- Jan 17, 2013

- 1,667

- Thread starter
- #5

- Feb 21, 2013

- 739

Yeah I mean domain, Thats what the problem is asking for ? Sorry I did not know it's called domain in english

- Jan 17, 2013

- 1,667

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

- Thread starter
- #7

- Feb 21, 2013

- 739

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?

- Admin
- #8

- Jan 26, 2012

- 4,042

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. \(\displaystyle f \left( \sqrt{\frac{8}{3}} \right), \hspace{1 mm} f(0), \hspace{1 mm} f \left( \frac{1}{4} \right), \hspace{1 mm} f(2)\)

- Thread starter
- #9

- Feb 21, 2013

- 739

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?

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?

Last edited:

- Admin
- #10

- Jan 26, 2012

- 4,042

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

- Thread starter
- #11

- Feb 21, 2013

- 739

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

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

- Admin
- #12

- Jan 26, 2012

- 4,042

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

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:\(\displaystyle \forall x>\frac{1}{4} \left(\frac{1}{4},2 \right)\)

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

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