Who can solve this problem???

C

CuongChi

New member
Joined
Dec 5, 2019
Messages
22
f(x)=2x^4-4x^3+3mx^2-mx-2m.sqrt(x^2-x+1)+2. Find m such that f(x)>=0 all x in real
 
Do you know any calculus (such as finding the minimum of this function)?

Please follow the guidelines by showing some work or telling us where you are stuck.
 
Dr.Peterson said:
Do you know any calculus (such as finding the minimum of this function)?

Please follow the guidelines by showing some work or telling us where you are stuck.
Click to expand...
Yes, I have know a little calculus. My idea is let m in a left side and function of x in the other side, but it's difficult for me to calculate, so I need some hind. Apologise for my English, i don't write it well
 
Find f' (x) and then try to solve f'(x)=0. Please show us your work.
 
CuongChi said:
i know that f(x)>=0=f(1) so f'(1)=0 and figure out m=2, but i was wondering if m could be a only value
Click to expand...
To be honest I am getting tired of looking at your posts and not seeing you show any work. You were told what to do but you refuse to do what we ask. Either answer our questions or go away. Personally I prefer you stay around but you have to play by our rules.
 
f'(x)=8x^3-12x^2+6mx-m-m(2x-1)/sqrt(x^2-x+1), indeed i don't know how to solve this
 
CuongChi said:
f'(x)=8x^3-12x^2+6mx-m-m(2x-1)/sqrt(x^2-x+1), indeed i don't know how to solve this
Click to expand...
Set that derivative to 0 then bring the square root to the other side and square both sides.
Let's see where you can go from there.
 
Last edited:
CuongChi said:
f'(x)=8x^3-12x^2+6mx-m-m(2x-1)/sqrt(x^2-x+1), indeed i don't know how to solve this
Click to expand...
Assuming your calculation above is correct:

0 = 8x^3 - 12x^2 + 6mx - m - m(2x-1) / sqrt(x^2-x+1)

m * (2x - 1) / sqrt(x^2-x+1) = 8x^3 - 12x^2 + 6mx - m

Now square both sides to eliminate the 'sqrt' part...... and continue....
 
Subhotosh Khan said:
Assuming your calculation above is correct:

0 = 8x^3 - 12x^2 + 6mx - m - m(2x-1) / sqrt(x^2-x+1)

m * (2x - 1) / sqrt(x^2-x+1) = 8x^3 - 12x^2 + 6mx - m

Now square both sides to eliminate the 'sqrt' part...... and continue....
Click to expand...
No, I think It's hard to calculate
 
Subhotosh Khan said:
Assuming your calculation above is correct:

0 = 8x^3 - 12x^2 + 6mx - m - m(2x-1) / sqrt(x^2-x+1)

m * (2x - 1) / sqrt(x^2-x+1) = 8x^3 - 12x^2 + 6mx - m

Now square both sides to eliminate the 'sqrt' part...... and continue....
Click to expand...
What do you get when you square both sides?
 
-36m^2.x^4+48m^2.x^3-45m^2.x^2+9m^2x-96mx^6+256mx^5-280mx^4+184mx^3-24mx^2-64x^8+256x^7-400x^6+336x^5-144x^4=0
 
You must log in or register to reply here.
Top