This is the question at which I'm stuck. (no grammar errors, right? Good: no grammar nazis...)

Let f(x) be polynomial and 3f(x) + 5f(-x) =x^{2 }+ 7. Find f(x).

I haven't used this site in a reaaal long time so I forgot how to insert the characters.... And, I'm transating this question from another language. So, please forgive me on any gramatical, format or spelling errors.

Thanks~

Guest Jan 6, 2016

#18**+10 **

I'd like to look at this one in some detail.......I wouldn't have known how to approach it without "guest's" suggestion......so.....I have to give him/her the lion's share of credit.....!!!!

3f(x) + 5f(-x) =x^2 + 7 find f(x)

Let f(x) be of the form ax^2 + bx + c

So

3f(x) = 3ax^2 + 3bx + 3c and let 5f(-x) = 5ax^2 - 5bx + 5c

And

[3ax^2 + 3bx + 3c ] + [5ax^2 - 5bx + 5c] = x^2 + 7

So, equating coefficients.....we have

[5a + 3a] = 1 → 8a = 1 → a = (1/8) and

[3b - 5b] = 0 → b = 0

[3c + 5c ] = 7 → 8c = 7 → c = 7/8

So.......

f(x) = (1/8)x^2 + 7/8

Proof:

3f(x) = (3/8)x^2 + 21/8 and 5f(-x) = (5/8)x^2 + 35/8

And adding these, we have........ x^2 + 56/8 = x^2 + 7

I liked this one........thanks, guest......!!!

I leran something new on here almost every day !!!!

CPhill
Jan 8, 2016

#4**0 **

3f(x) + 5f(-x) =x2 + 7. Find f(x).

Oh! I just re-read it .. . the "3" I'm speaking of is part of the equation :/

Hayley1
Jan 6, 2016

#9**0 **

ROFL, yes, or she could be a dog who dresses up like a cat? I forgot the exact analogy he/she used:(

Coldplay
Jan 6, 2016

#14**0 **

Well you got that wrong Miss Smartypants Coldplay!

I was not admiring my lantern post.

EVEN though it is indeed a very fine lantern post. :D

I was admiring myself in my looking glass - So there

Melody
Jan 6, 2016

#16**+5 **

Let f(x) be polynomial and 3f(x) + 5f(-x) =x2 + 7. Find f(x).

Mmm tricky...

No sorry it is all too much for me.

http://www.wolframalpha.com/input/?i=3f%28x%29%2B5f%28-x%29%3Dx%5E2%2B7

Melody
Jan 6, 2016

#17**+5 **

Let

\(\displaystyle f(x) = ax^{2}+bx+c\) ,

substitute this into the lhs of the equation and equate coefficients.

Guest Jan 6, 2016

#18**+10 **

Best Answer

I'd like to look at this one in some detail.......I wouldn't have known how to approach it without "guest's" suggestion......so.....I have to give him/her the lion's share of credit.....!!!!

3f(x) + 5f(-x) =x^2 + 7 find f(x)

Let f(x) be of the form ax^2 + bx + c

So

3f(x) = 3ax^2 + 3bx + 3c and let 5f(-x) = 5ax^2 - 5bx + 5c

And

[3ax^2 + 3bx + 3c ] + [5ax^2 - 5bx + 5c] = x^2 + 7

So, equating coefficients.....we have

[5a + 3a] = 1 → 8a = 1 → a = (1/8) and

[3b - 5b] = 0 → b = 0

[3c + 5c ] = 7 → 8c = 7 → c = 7/8

So.......

f(x) = (1/8)x^2 + 7/8

Proof:

3f(x) = (3/8)x^2 + 21/8 and 5f(-x) = (5/8)x^2 + 35/8

And adding these, we have........ x^2 + 56/8 = x^2 + 7

I liked this one........thanks, guest......!!!

I leran something new on here almost every day !!!!

CPhill
Jan 8, 2016