Can You Find f(0)?
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- Опубліковано 1 гру 2023
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This function is impossible. If x=0, then f(0-0-3)=f(-3)=0-7*0=0. If x=1, then f(1-1-3)=f(-3)=1-7*1=-6. So, f(-3)=0 and f(-3)=-6 at the same time. Such function cannot exist. This exercise is incorrect.
You could say it’s not a function, or you could say that there are two functions that satisfy this equation. For both functions f(0) has the same value.
@@SWalkerTTU if it is a function, so what's the value of f(-3)? It must be only one number. But it can be 0 and -6 at the same time. It is impossible for the function.
The exercise is correct.
The functional equation has 2 solutions, which may be expressed as a function with one two valued parameter (k):
f(x ; k) = (x + 3) (x + 4) + x (1 + k √(4x + 13)) , (k = ± 1)
This function satisfies the original functional equation and: f(-3 ; 1) = - 6 ; f(-3 ; -1) = 0
Moreover, f(0 ; 1) = f(0 ; -1) = 12
Could be a multivalued function, tho that's by definition not a function
@@user-bw7yp9ji4u The thing being discussed is not a function, but an equation. There are two functions that satisfy it, and they only have the same value for 0 and -13/4.
We want to find f(0), so we have the requirement
x^2 - x - 3 = 0 or
x^2 = x + 3.
Thus we get
x^4 = (x + 3)^2 = x^2 + 6x + 9
and
x^4 - 7x = x^2 + 6x + 9 - 7x = x^2 - x + 9
Replace x^2 by x + 3 once again:
= x + 3 - x + 9
= 12.
Altogether, we have x^2 - x - 3 = 0, so for these special values of x, we get
f(0) = f(x^2 - x - 3) = x^4 - 7x = 12.
So twelve is the final result.
I prefer the second method which is very sneaky and less messy!
Thanks for the feedback, Robert! 🙂
Yeah, right when he started to work through the second method I felt there was going to be some way to manipulate the expression inside the argument to look like what was on the right of the equal sign.
Briliant!
Thank you
This is really just an elementary problem about polynomials. If we have f(0) = k, then we must have x⁴ − 7x = k or x⁴ − 7x − k = 0 for both values of x which satisfy x² − x − 3 = 0. So, x² − x − 3 must be a _factor_ of x⁴ − 7x − k which implies that k will be the _remainder_ of (x⁴ − 7x)/(x² − x − 3). Therefore, all that is required is a polynomial division, and if we do that we find that
x⁴ − 7x = (x² + x + 4)(x² − x − 3) + 12
so k = 12.
I wanted to post more or less the same. Many of the problems posted on this channel can be solved by using polynomlial division at some point. It is simpler because it doesn't require any guesswork as how to factor certain polynomials even if sometimes it is more tideous.
I’m not sure if this is valid or not. Initially I solved for X, but didn’t want to go through the task of raising it to the 4th power, so I tried the following. I found an expression that I could multiply X^2-X-3 by to get X^4-7X-12. That expression is X^2+X+4. I then had to add 12 to get rid of the -12 by adding 12. So, if X^2-X-3 is 0, then 0*(X^2+X+4) + 12=12.
I did the hard work and set x²-x-3 to y then solved for x and substituted the RHS with it. Boy that was no fun!
So what did I get? f(y) = y² + 8y + 12 ± y √(13 + 4y). So there are always two possible function values except for y = 0 and y = -13 / 4.
The original functional equation has 2 solutions, which may be expressed as a function with one two valued parameter (k):
f(x ; k) = (x + 3) (x + 4) + x (1 + k √(4x + 13)) , (k = ± 1)
Moreover, f(0 ; 1) = f(0 ; -1) = 12
The second method reminds me of Galois fields.
I found a glitch with f(x).
In original equation replace x with zero. f(-3)=0
In the same equation replace x with 1 then f(-3)=-6
So f(x) violates the basic properties of a function.
To make f(x) a proper function the question may be modified to f(x^2-x-3)=x^4-2x^3+x+18
This way for both values of x =1 or0 f(-3) =18
F(x)=x^2+5x+12
And f(0)=12. !!
How do you like that??
This problem can be viewed as x^2-x-3=0 ---- (1) and find x^4-7x --- (2).
We can then lessen the degree of (2) by rewriting (1) to x^2=x+3. 🤗🤗🤗🤗🤗🤗
f(x²-x-3)=x⁴-7x, find f(0)
There 2 ways to solve the problem:
• f(0)=f(x|x²-x-3=0)
In another words we have to find f(x) if x²-x-3=0 then plug the roots to RHS. As it is a qudratic equation there are 2 roots, and even worst if the roots are not integers as the presence of x⁴ in RHS
• Manipulate RHS and noting that x²-x-3=0 as follows:
x⁴-7x=x(x³-7)
=x[(x²-x-3)x+x²+3x-7]
=x(x²+3x-7) as x²-x-3=0
=x[(x²-x-3)+4x-4]
=4(x²-x)
=4(x²-x-3)+12
=12
Therefore f(0)=12
The second method is faster !
Oh man, functional equations always break my brain. I guess I need to practice more to get better...
Absolutely! The more you do the better
@@SyberMathPractice is good to keep you functional 👍
If you replace x=0, then f(-3)= 0, but if you replace by x=-1, you will get f(-3) =8, so f(x) is not an uniform function. Are there something wrong in my reasoning?
Apart from it being completely irrelevant to the question, no.
To find f(x) let's suppose x^2-3=0. Then replace x^4 by 9, we get f(-x)=9-7x and finally by replacing x by -x we get f(x)=7x+9
Guzel soru hocam eline saglik
Saol varol!
@@SyberMath: So now I caught you as I did many many months before that implied you are a teacher and now trying to figure out for a long time your accent with this Turkish language response telling us your are probably of Turkish origin!
@@roberttelarket4934 😁
x^4-7x divided by x^2-x-3 has the residual 12. 2 mins.
3rd method! 👏🏻
I also got 12 as the answer.
Wow
😍
so easy
f(x)=x^2+8x+12+x√(13+4x)
Also …-x sqrt(4x + 13).
@@SWalkerTTUyeah, that’s what I got. I wonder if there is a way to eliminate one of the possibilities or if we just accept that there are two possibilities.
@@GreenMeansGOF There simply are two function solutions to this equation, but since f(0) for both functions is the same, it doesn’t matter which you choose. f(-13/4) is also the same for both functions. This particular question is answerable only because the poser chose a value for evaluation that gave a single result.