If 4f(x) + f(3-x) = x^2 what is f(x)? Can somebody guide me through this?
Subtract f(3-x) from both sides and then divide both sides by 4.
hmmm
so... it would be f(x)= \[x^{2}-(3-x)/4\]
i think the gimmick is to replace x by 3 - x in the first equation and get \[4f(3-x)+f(x)=(3-x)^2\] and then you have two equations \[4f(x)+f(3-x)=x^2\] \[4f(3-x)+f(x)=(3-x)^2\] then some algebra to solve for \[f(x)\] if i recall a system of two equations. let me write it out on paper but i am pretty sure this is the method you need
wow. that helps a lot. I wasnt sure if i could just subtract 3-x. It was part of the function.
Multiply the first equation by four Minus the first one by the second one Isolate f(x)
it is cumbersome to write, so lets put \[a=f(x),b=f(3-x)\] and solve \[4a+b=x^2\] \[4b+a=x^2-6x+9\] or the system \[4a+b=x^2\] \[a+4b=x^2-6x+9\] we want "a" so lets multiply the first equation by 4 and get \[8a+4b=4x^2\] subtract the second equation from it and get \[3a=3x^2+6x-9\] divide by 3 and get \[a=x^2+2x-3\] so \[f(x)=x^2+2x-3\] check my algebra
well that was wrong. when i subtract i should have \[7a=3x^2+6x-9\] making \[a=\frac{3x^2+6x-9}{7}\] first one i made a mistake
But if a = f(x), it doesn't check in the original equation.
Actually it should be a =( x^2+2x-3)/5
so f(x) = (x^2 + 2x -3) /5?
Yes. Check is out by using various x values.
where did the 5 come from?
-15 f(x) = -4x^2+x^2 -6x+9
That was the result after eliminating 4f(3-x) using elimination.
oh of course! i is true that \[4\times 4=16\] not for example 8 like i wrote
so we should have \[15a=15f(x)=3x^2+6x-9\] and so \[f(x)=\frac{x^2+2x-3}{5}\] damn that arithmetic
Ain't it the truth!!
and the check is good! http://www.wolframalpha.com/input/?i=4%28x^2%2B2x-3%29%2F5%2B%28%283-x%29^2%2B2%283-x%29-3%29%2F5%29
Hey guys, thanks so much. Can you help me determine the total revenue function of another problem?
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