Solving x^4 - x^3 + x - 1 = 0 How is this done?

  • Thread starter Thread starter sdoman
  • Start date Start date
AI Thread Summary
To solve the equation x^4 - x^3 + x - 1 = 0, the discussion focuses on differentiating the function F(x) = e^(2x)/x^4 using the Quotient Rule. The differentiation leads to the expression 2x^4(e^(2x)) - 4x^3(e^(2x))/x^8. Participants clarify that while e^(2x) does not cancel, it can be factored out, aiding in simplification. The final answer is presented as 2e^(2x)(x-2)/x^5, emphasizing the importance of correctly managing the factors during simplification. Understanding these steps is crucial for successfully differentiating the function.
sdoman
Messages
6
Reaction score
0
How is this done?

x^4 - x^3
 
Physics news on Phys.org
What do you need to do?

You deleted the three sections, the first of which shows the complete problem and variables and data.
 
The original problem is to differentiate the following:

F(x) =

e^2x
x^4

So I applied the Quotient Rule and have it simplified to

2x^4(e^2x)-4x^3(e^2x)
x^8

I see that the e^2x's cancel and I don't understand how to simplify the rest.


The answer is
2e^2x(x-2)
x^5

I just can't put the pieces together, thanks!
 
Youre almost there. Just pull out some x's and the exponential, like, x^3 e^2x (2 x^1 - 4)/x^8.
Then cancel out those x^3 with those on the bottom, and pull out another two and you have the answer.
 
sdoman said:
The original problem is to differentiate the following:

F(x) =

e^2x
x^4

So I applied the Quotient Rule and have it simplified to

2x^4(e^2x)-4x^3(e^2x)
x^8
This (above) is correct.
sdoman said:
I see that the e^2x's cancel and I don't understand how to simplify the rest.
No, the e^(2x)'s don't cancel out. You can factor them out of each term, though.
sdoman said:
The answer is
2e^2x(x-2)
x^5

I just can't put the pieces together, thanks!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top