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) =x2 + 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~
+130466 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 !!!!
+8581 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 :/
+4084 ROFL, yes, or she could be a dog who dresses up like a cat? I forgot the exact analogy he/she used:(
+118696 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 
+118696 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
Let
f(x)=ax2+bx+c ,
substitute this into the lhs of the equation and equate coefficients.
+130466 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 !!!!
