Question

3f(x-2) + f(2-x) = 5x
f(x) = ?

Answer 1

\[\text{ is \it } f(x)=\frac{5}{2}(x+2)\]
checking..
one sec

Answer 2

thats super fast xD how did you..

Answer 3

well the sum of odd functions should be odd

Answer 4

f(2-x)=f(-(x-2))=-f(x-2)

Answer 5

^_^ *

Answer 6

\[2f(x-2)=5x\]

Answer 7

\[f(x-2)=\frac{5x}{2} \\ \text{ \let } u=x-2 \\ \text{ then } u+2=x \\ f(u)=\frac{5}{2}(u+2)\]

Answer 8

checking:
\[3f(x-2)+f(2-x) \\ =3\frac{5}{2}(x-2+2)+\frac{5}{2}(2-x+2) \\ =\frac{15}{2}(x)+\frac{5}{2}(4)-\frac{5}{2}x \\ =\frac{10}{2}x+10 \\ =5x+10\]
but I think I made a mistake

Answer 9

i misplaced the solution, one sec...

Answer 10

I guess odd sum doesn't have to mean both addends are odd

Answer 11

Well I was just about to ask lol.

Answer 12

I see.. but that looks new and interesting... never thought about doing that before xD

Answer 13

so the question is still open..

Answer 14

I might have a solution let me finish solving this matrix...

Answer 15

okie :) for solving, i think we can assume f(x) is continuous/differentiable/smooth/good and a nice function...

Answer 16

let's do the sensible thing here and x-2=y so we have: \[3f(y)+f(-y)=5(y+2)\] Here's my trick, exploit the fact that every function is the sum of an even and odd function: \[f(x) = \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}=h(x)+g(x)\] h(x) is that first, even one, and g(x) is the second, odd one, pretty straight forward so far I hope? Ok now let's substitute in f(x)=h(x)+g(x) to get: \[3h(y)+3g(y)+h(-y)+g(-y)=5(y+2)\] use the properties of even and odd functions to get the first equation below and then let y be -y to get the second equation: \[4h(y)+2g(y)=5(y+2) \\ 4h(y)-2g(y)=5(-y+2)\] I don't know why I used a matrix to solve this, when it's so obvious to solve by adding and subtracting the two equations, to get: \[8h(y)=20 \\ 4g(y) = 10y\] Now plug them in since f(x)=h(x)+g(x)... Also I just realized I don't know my alphabet, it should be fgh not fhg... uh... whoops lol.

Answer 17

O.O

Answer 18

Is f(x) a linear function? if so you can begin the set up with f(x) = mx +b
then we can solve for m and b

Answer 19

it could be linear or exponential, we won't know until we solve it..
im still going through Kainui's solution xD

Answer 20

I'm getting that f(x)=5/2(x-1) I think, unless I made some algebra mistake, I'm checking to make sure it meets the question's requirements.

Answer 21

I'm checking it too

Answer 22

\[8h(y)=20 \\ 4g(y) = 10y\]

gives \[f(x) = \frac{5}{2}(x+1)\]

right ?

Answer 23

and if so it is correct
awesome work @Kainui

Answer 24

When I plugged that into the initial formula this didn't work. What did you get?
For example 3f(x-2) will become 3(5/2((x-2)-1) right?

Answer 25

and f(2-x) is 5/2((2-x)-1)

Answer 26

if we add them we need 5x

Answer 27

\[3 \cdot \frac{5}{2}(1+x-2)+\frac{5}{2}(1+2-x) \\ \frac{15}{2}(x-1)+\frac{5}{2}(3-x) \\ \frac{15}{2}x-\frac{15}{2}+\frac{15}{2}-\frac{5}{2}x \\ \frac{10}{2}x \\ 5x\]

Answer 28

this f(x) = odd + even is really a neat thing !

Answer 29

Yeah I don't know how I figured that out, I just got lucky.

Answer 30

But it might be a pretty general method for solving these things.

Answer 31

also @freckles I don't know how you got the algebra to work out when me and mira couldn't cause I'm still struggling somewhere.

Answer 32

i had a solution using taylor series but im bit embarassed now to post after seeing that cute method lol

Answer 33

@Kainui did you use the corrected function?

Answer 34

Kainui sometimes I think you have a portion of Einstein's intelligence implemented somewhere in your brain

Answer 35

\[f(x) = \frac{5}{2}(x+1) \]

Answer 36

@ganeshie8 Wait show it! I was trying to find the taylor series type solution with some kind of recursion but I couldn't so that's why I did this junk! XD

Answer 37

Hahaha thanks @Miracrown !!! I wish I could put that on my resume right now.

Answer 38

don't be embarrassed
I posted something totally wrong
and I don't care

Answer 39

Kainui you can, there is copy and paste for a reason

Answer 40

also wasn't trying to imply what you are about to do is wrong
i was saying who can be more embarrassed than me
and i'm not embarrassed so you shouldn't be either :p

Answer 41

lol freckles let me remind kainui of this quote: ''Be who you are and say what you feel, because those who mind don't matter and those who matter don't mind''

Answer 42

\(3f(x-2) + f(2-x) = 5x\)

plugin \(x = 2\), we get \(f(0) = 5/2\)

differentiating we get : \(3f'(x-2) -f'(2-x) = 5 \implies f'(0) = 5/2\) and \(f^n(0) = 0\) for \(n\gt 1\)


\(\cdots\)

Answer 43

nope figured it out

Answer 44

you used that little equation there

Answer 45

Here's what I did and got a similar result:

3f(x-2) + f(2-x) = 5x

F(x) = mx+b

3m(x-2) + 3b+m(2-x)+b=5x

3mx-6m+3b+2m-mx+b=5x

2mx=5x and 4b-4m = 0

m = 5/2
b = 5/2

f= 5/2 (x+1)

Answer 46

i think that works perfectly too as it gives the function that satisfies the given equation xD

freckels :
\(3f'(x-2) -f'(2-x) = 5~; ~\color{Red}{\text{ plugin x = 2}~}\implies f'(0) = 5/2 \)

Answer 47

Oh you figured it out already okieee

Answer 48

I like your method @ganeshie8 and @Miracrown

Answer 49

and @Kainui
god don't want to leave anyone out

Answer 50

or gosh if you prefer gosh

Answer 51

lol i wish to remember writing a function as "odd+even" while solving other problems xD

Answer 52

That's a good approach, we just need to make sure we get the right answer when using it

Answer 53

\(\color{blue}{\text{Originally Posted by}}\) @freckles
or gosh if you prefer gosh
\(\color{blue}{\text{End of Quote}}\)
I prefer Jambalaya actually...

Answer 54

ich bin gott!!!

Answer 55

lol

Answer 56

Here's another thing I considered using that's sort of similar to the power series method @ganeshie8 the derivative of an even function is an odd, and odd is even: \[\large \left[ \frac{f(x)+f(-x)}{2} \right]' = \frac{g(x)-g(-x)}{2}\]