If 3f(x)+f(3-x)=x squared, what's f(x)?

[pugfug90]

Homework Statement

Homework Equations

The Attempt at a Solution

I first tried replacing "x" when in parenthese with y. Therefor, 3y+3-y=x, yadda yadda, 2y=x-3, y=(x-3)/2, however, the real answer turned out to be some weird fractional thing. Please give me some kind of direction..

EDIT: Ignore that...

 

[nrqed]

I first tried replacing "x" when in parenthese with y. Therefor, 3y+3-y=x, yadda yadda, 2y=x-3, y=(x-3)/2, however, the real answer turned out to be some weird fractional thing. Please give me some kind of direction..

OTHER EDIT!
Sorry! yes, my initial trick does work if you replace x by 3-x. I am obviously not thinking straight after having been teaching for 4 hours straight! just replace x by 3-x and then isolate your f(x)!


Ignore the rest...



EDIT: oops. I had read too quickly. My trick won't work right away. You will also have to consider replacing x by minus x to make it work

Start from the initial equation. Substitute x -> x-3 everywhere you see an x (on both sides of the equation. This will give you a new equation containing f(x) and f(x-3). Now use those two equations to get rid of f(3-x) to leave you with an answer for f(x)

 

[pugfug90]

3f(3-x)+f(x)=3-x is what I get if I plug in 3-x where x is...

 

[nrqed]

3f(3-x)+f(x)=3-x is what I get if I plug in 3-x where x is...

you mean (3-x)^2 on the right side??

 

[pugfug90]

Oops haha
So 3f(3-x)+f(x)=(x squared)+9-6x.

So can you tell me where to go from here? I tried substituting y into x again, get (x squared - 6x)/-2, while the real answer is something like x/3+x/5-3 or something. Also, where did the 'plug 3-x for x everywhere' thing came from? And if this is a "topic" I can look up like for quadratic equations, or "f and g compositions"?

 

[nrqed]

Oops haha
So 3f(3-x)+f(x)=(x squared)+9-6x.

So can you tell me where to go from here? I tried substituting y into x again, get (x squared - 6x)/-2, while the real answer is something like x/3+x/5-3 or something. Also, where did the 'plug 3-x for x everywhere' thing came from? And if this is a "topic" I can look up like for quadratic equations, or "f and g compositions"?

if you substitute again you will get back to the initial equation. No, just use the two equations you now have. get rid of f(3-x) (isolate it from one equation and plug in the other one) and you can isolate f(x) now.

 

[qspeechc]

I am sorry, I was just reading through this thread: how can you substitute x for x-3? Doesn't this imply -3=0?

 

[Hurkyl]

I am sorry, I was just reading through this thread: how can you substitute x for x-3? Doesn't this imply -3=0?

He said substitute for, not set equal to.

If it makes you feel better, try the (equivalent) substitution x -> y - 3.

 

[pugfug90]

3f(3-x)+f(x)=(x squared)+9-6x

get rid of f(3-x) (isolate it from one equation and plug in the other one) and you can isolate f(x) now.
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If I want to isolate f(x).. it'd be f(x)=xsquared+9-6x-3f(3-x), right?