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Mathematics 7 Online
OpenStudy (anonymous):

If 4f(x) + f(3-x) = x^2 what is f(x)? Can somebody guide me through this?

OpenStudy (mertsj):

Subtract f(3-x) from both sides and then divide both sides by 4.

OpenStudy (anonymous):

hmmm

OpenStudy (anonymous):

so... it would be f(x)= \[x^{2}-(3-x)/4\]

OpenStudy (anonymous):

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

OpenStudy (anonymous):

wow. that helps a lot. I wasnt sure if i could just subtract 3-x. It was part of the function.

OpenStudy (anonymous):

Multiply the first equation by four Minus the first one by the second one Isolate f(x)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (mertsj):

But if a = f(x), it doesn't check in the original equation.

OpenStudy (mertsj):

Actually it should be a =( x^2+2x-3)/5

OpenStudy (anonymous):

so f(x) = (x^2 + 2x -3) /5?

OpenStudy (mertsj):

Yes. Check is out by using various x values.

OpenStudy (anonymous):

where did the 5 come from?

OpenStudy (mertsj):

-15 f(x) = -4x^2+x^2 -6x+9

OpenStudy (mertsj):

That was the result after eliminating 4f(3-x) using elimination.

OpenStudy (anonymous):

oh of course! i is true that \[4\times 4=16\] not for example 8 like i wrote

OpenStudy (anonymous):

so we should have \[15a=15f(x)=3x^2+6x-9\] and so \[f(x)=\frac{x^2+2x-3}{5}\] damn that arithmetic

OpenStudy (mertsj):

Ain't it the truth!!

OpenStudy (anonymous):

Hey guys, thanks so much. Can you help me determine the total revenue function of another problem?

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