if 4f(x)+f(2-x)=x^2 , what is f(x)?
- Thread starter jburke15
- Start date
D
Deleted member 4993
Guest
jburke15 said:
Please share with us your work, indicating exactly where you are stuck, so that we know where to begin to help you.
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,204
\(\displaystyle 4f(x) + f(2-x) = x^2\)
\(\displaystyle 4f(2-x) + f[2 - (2-x)] = (2-x)^2\)
\(\displaystyle 4f(2-x) + f(x) = (2-x)^2\)
\(\displaystyle f(x) + 4f(2-x) = (2-x)^2\)
multiply the original equation by -4 and add to the last equation above ...
\(\displaystyle -16f(x) - 4f(2-x) = -4x^2\)
\(\displaystyle f(x) + 4f(2-x) = (2-x)^2\)
----------------------------------------------
\(\displaystyle -15f(x) = (2-x)^2 - 4x^2\)
solve the last equation for f(x).
\(\displaystyle 4f(2-x) + f[2 - (2-x)] = (2-x)^2\)
\(\displaystyle 4f(2-x) + f(x) = (2-x)^2\)
\(\displaystyle f(x) + 4f(2-x) = (2-x)^2\)
multiply the original equation by -4 and add to the last equation above ...
\(\displaystyle -16f(x) - 4f(2-x) = -4x^2\)
\(\displaystyle f(x) + 4f(2-x) = (2-x)^2\)
----------------------------------------------
\(\displaystyle -15f(x) = (2-x)^2 - 4x^2\)
solve the last equation for f(x).