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 !!!!
