Finding the function that will satisfy the equation.

**ernestjohn** · August 24th 2012, 04:16 PM

If g(x) satisfies 4g(x) + g(1-x) = 3x², what is what is g(x)?

I'm really stuck, what is a possible solution?

**Prove It** · August 24th 2012, 05:13 PM

I expect since this particular combination of functions gives you a quadratic, you will probably find that g(x) is a quadratic. So let $\displaystyle \begin{align*} g(x) = a\,x^2 + b\,x + c \end{align*}$ , then $\displaystyle \begin{align*} g(1 - x) = a(1 - x)^2 + b(1 - x) + c = a - 2a\,x + a\,x^2 + b - b\,x + c \end{align*}$ . So

$\displaystyle \begin{align*} 4g(x) + g(1 - x) &= 3x^2 \\ 4a\,x^2 + 4b\,x + 4c + a - 2a\,x + a\,x^2 + b - b\,x + c &= 3x^2 \\ 5a\,x^2 + (3b - 2a)x + a + b + 5c &= 3x^2 + 0x + 0 \end{align*}$

So this means $\displaystyle \begin{align*} 5a = 3 \implies a = \frac{3}{5}, 3b - 2a = 0 \implies b = \frac{2}{5}, a + b + 5c &= 0 \implies c = -\frac{1}{5} \end{align*}$ .

Therefore a function which satisfies your original equation is $\displaystyle \begin{align*} g(x) = \frac{3}{5}x^2 + \frac{2}{5}x - \frac{1}{5} \end{align*}$ .

**MaxJasper** · August 24th 2012, 05:52 PM

Simple solution:

Substitute $x\to 1-x$ ...you get a second equation, eliminate g(1-x) between them to obtain g(x):

$g(x)\text{:=}\frac{1}{15} \left(9 x^2+2 x-3\right)$

**richard1234** · August 24th 2012, 09:55 PM

Replace $x$ with $1-x$ to obtain

$4g(1-x) + g(x) = 3(1-x)^2$ . We already know

$4g(x) + g(1-x) = 3x^2$

Multiply the second equation by -4 and add them.

$-15g(x) = 3(1-x)^2 - 12x^2$

Divide both sides by -15 and simplify.

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