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# Something about functions or something...

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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+ 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~

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 !!!!   Jan 8, 2016

#1
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Jan 6, 2016
edited by Guest  Jan 6, 2016
#2
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Wouldn't you plug in 3 for the variables? :)

Jan 6, 2016
#4
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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 :/

Jan 6, 2016
#5
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I'll get Mrs. Melody :)

Jan 6, 2016
#7
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Lol, Haley did you forget what I told you? It's Mr.Melody!

Jan 6, 2016
#8
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MR?!

Jan 6, 2016
#9
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ROFL, yes, or she could be a dog who dresses up like a cat? I forgot the exact analogy he/she used:(

Jan 6, 2016
#10
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She is on her way! :)

Jan 6, 2016
#11
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Hopefully, Maybe.. :)

Jan 6, 2016
#12
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LOL, she's probably admiring her lantern post.

Jan 6, 2016
#13
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What?!?!?! MR MELODY?? MR?? (Oops, no offence)

Jan 6, 2016
#14
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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  Jan 6, 2016
#15
+10

ColdPlay, I do believe that Mrs. Melody just roasted you (:

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

Jan 6, 2016
#17
+5

Let

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

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

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
#19
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Plz stop callin me smarty pants Melody, LOL!

And Haley, I was not roasted, I was merely sautéed...

Jan 10, 2016
#20
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How did you find the f(x) to substitute in the first place?

Jan 26, 2016