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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

Best Answer 

 #18
avatar+104793 
+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 !!!! 

 

 

 

 

cool cool cool

 Jan 8, 2016
 #1
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0

Melody? radix? Someone?

 Jan 6, 2016
edited by Guest  Jan 6, 2016
 #2
avatar+8613 
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Wouldn't you plug in 3 for the variables? :)

 Jan 6, 2016
 #3
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0

What do you mean?

 Jan 6, 2016
 #4
avatar+8613 
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 :/

 Jan 6, 2016
 #5
avatar+8613 
0

I'll get Mrs. Melody :)

One moment please!

 Jan 6, 2016
 #6
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Ok, thank you~~

 Jan 6, 2016
 #7
avatar+4080 
0

Lol, Haley did you forget what I told you? It's Mr.Melody!

 Jan 6, 2016
 #8
avatar+8613 
0

MR?!

 Jan 6, 2016
 #9
avatar+4080 
0

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
avatar+8613 
0

She is on her way! :)

 Jan 6, 2016
 #11
avatar+8613 
0

Hopefully, Maybe.. :)

 Jan 6, 2016
 #12
avatar+4080 
0

LOL, she's probably admiring her lantern post.

 Jan 6, 2016
 #13
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0

What?!?!?! MR MELODY?? MR?? (Oops, no offence) 

 Jan 6, 2016
 #14
avatar+105604 
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  cheeky

 

 

 Jan 6, 2016
 #15
avatar+8613 
+10

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

 Jan 6, 2016
 #16
avatar+105604 
+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
avatar
+5

Let

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

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

 Jan 6, 2016
 #18
avatar+104793 
+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 !!!! 

 

 

 

 

cool cool cool

CPhill Jan 8, 2016
 #19
avatar+4080 
0

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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0

How did you find the f(x) to substitute in the first place?

 Jan 26, 2016

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