+0

# Something about functions or something...

+5
349
20

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~

Guest Jan 6, 2016

#18
+78741
+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
Sort:

#1
0

Guest Jan 6, 2016
edited by Guest  Jan 6, 2016
#2
+8621
0

Wouldn't you plug in 3 for the variables? :)

Hayley1  Jan 6, 2016
#3
0

What do you mean?

Guest Jan 6, 2016
#4
+8621
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
#5
+8621
0

I'll get Mrs. Melody :)

Hayley1  Jan 6, 2016
#6
0

Ok, thank you~~

Guest Jan 6, 2016
#7
+4098
0

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

Coldplay  Jan 6, 2016
#8
+8621
0

MR?!

Hayley1  Jan 6, 2016
#9
+4098
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
#10
+8621
0

She is on her way! :)

Hayley1  Jan 6, 2016
#11
+8621
0

Hopefully, Maybe.. :)

Hayley1  Jan 6, 2016
#12
+4098
0

LOL, she's probably admiring her lantern post.

Coldplay  Jan 6, 2016
#13
0

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

Guest Jan 6, 2016
#14
+91045
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
#15
+8621
+10

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

Hayley1  Jan 6, 2016
#16
+91045
+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
+78741
+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
+4098
0

Plz stop callin me smarty pants Melody, LOL!

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

Coldplay  Jan 10, 2016
#20
0

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

Guest Jan 26, 2016

### 6 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details