MHB How to Find Max and Min Points for \(\frac{x^3}{2} - |1 - 4x|\) in \((0, 2)\)?

[Petrus]

Calculate max and min point to function $$\frac{x^3}{2}-|1-4x|$$ in the range $$\left(0,2 \right)$$
I got one question, shall I ignore when it's $$\frac{x^3}{2}-(-1+4x)$$ cause then $$x<0$$ and that don't fit in my range? Do I got correct?

 

[alyafey22]

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

 

[Petrus]

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

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

 

[alyafey22]

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

 

[alyafey22]

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

 

[Petrus]

let's make them to case 1 and case 2
Case 1:
$$|1-4x|=-(1-4x) \,\,\, \,\,\,\forall \,\,\, x>\frac{1}{4}$$ $$\left(\frac{1}{4},2 \right)$$
Our function become $$\frac{x^3}{2}-1+4x$$
If we derivate the function we get $$\frac{3x^2}{2}+4$$
Now we want to find critical point $$\frac{3x^2}{2}+4=0$$
So I factor out $$\frac{1}{2}$$ and get $$\frac{1}{2}(3x^2+8)=0$$ and our critical point (complex) are $$x_1=\sqrt{\frac{-8}{3}}$$ and $$x_2=-\sqrt{\frac{-8}{3}}$$ I guess we shall ignore case 1 cause we got complex roots.
Case 2:
$$|1-4x|=-(-1+4) \,\,\, \,\,\,\forall \,\,\ x<\frac{1}{4} $$ $$\left(0,\frac{1}{4} \right)$$
Our function become $$\frac{x^3}{2}+1-4x$$
If we derivate the function we get $$\frac{3x^2}{2}-4$$
Now we want to find critical point $$\frac{3x^2}{2}-4=0$$
So I factor out $$\frac{1}{2}$$ and get $$\frac{1}{2}(3x^2-8)=0$$ and our critical point are: $$x_1:\sqrt{\frac{8}{3}}$$ and $$x_2:-\sqrt{\frac{8}{3}}$$ but our $$x_2$$ 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, $$x=\sqrt{\frac{8}{3}}$$. Find the following values and then determine where you have a maximum and minimum. $$f \left( \sqrt{\frac{8}{3}} \right), \hspace{1 mm} f(0), \hspace{1 mm} f \left( \frac{1}{4} \right), \hspace{1 mm} f(2)$$

 

[Petrus]

Hello,
Thanks ZaidAlyafey and Jameson for the help! I did correct answer!:)(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: $$\frac{1}{128}$$
min: $$1-\frac{16\sqrt{\frac{2}{3}}}{3}$$
edit: what does $$\,\,\, \,\,\,\forall \,\,\,$$ 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 $$\,\,\, \,\,\,\forall \,\,\,$$ 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:

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

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

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