# [SOLVED]Max and point 2, absolute value

#### Petrus

##### Well-known member
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
$$\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}$$

#### Petrus

##### Well-known member
$$\displaystyle |1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}$$
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
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
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

##### Well-known member
MHB Math Helper
Ok , then you are correct , derive each function separately .

#### Petrus

##### Well-known member
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

Staff member
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
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?

Last edited:

#### Jameson

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

#### Petrus

##### Well-known member
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

$\forall$ means "for all" or "for every". Earlier you wrote 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) $$\displaystyle x \mid \frac{1}{4}\le x \le 2$$