We are given that [tex]4f(x)+f(5-x)=x^2[/tex].
Substitute x with 5-x, then the above equation becomes:
[tex]4f(5-x)+f(5-(5-x))=(5-x)^2[/tex], that is
[tex]4f(5-x)+f(x)=(5-x)^2[/tex]
So, we have the following system of equations:
i) [tex]4f(x)+f(5-x)=x^2[/tex]
ii) [tex]4f(5-x)+f(x)=(5-x)^2[/tex]
multiply the first equation by -4, so that we eliminate f(5-x)'s
i) [tex]-16f(x)-4f(5-x)=-4x^2[/tex]
ii) [tex]4f(5-x)+f(x)=(5-x)^2[/tex]
adding the 2 equations side by side we have:
[tex]-15f(x)=-4x^2+(5-x)^2[/tex]
expanding the binomial, and collecting same terms we have:
[tex]-15f(x)=-4x^2+(25-10x+x^2)[/tex]
[tex]-15f(x)=-3x^2-10x+25[/tex]
dividing by -5:
[tex]3f(x)=\frac{3}{5}x^2+2x-5[/tex]
dividing by 3:
[tex]\displaystyle{f(x)= \frac{1}{5}x^2+ \frac{2}{3}x-\frac{5}{3} [/tex]
Answer: [tex]\displaystyle{f(x)= \frac{1}{5}x^2+ \frac{2}{3}x-\frac{5}{3} [/tex]
