can you find f

There is no need to assume that $f$ is linear. Just plug in $-x$ to the original equation to get $\begin{aligned} f(x+1) + 2f(1-x) &= 3x-2 \\ f(-x+1) + 2f(1+x) &= -3x-2 \end{aligned}$ Multiplying the second equation by 2 and subtracting the first from the second gives $3f(1+x) = -9x-2,$ and setting $y = 1+x$ gives $\begin{aligned} 3f(y) &= -9(y-1)-2 \\ f(y) &= -3(y-1)-\frac23 = \frac73-3y. \end{aligned}$ And it's clear that $f(y) = \frac73-3y$ satisfies the original equation, so we are done.

Assuming $f(x)$ is a polynomial of $x$ , then $f(x)$ on the RHS has the same degree as the LHS which is 1. Then we can write $f(x)=ax+b$ , where $a$ and $b$ are constants. Abd we have:

$\begin{aligned} f(x+1) + 2f(1-x) & = 3x - 2 \\ a(x+1) + b + 2(a(1-x)+b) & = 3x - 2 \\ -ax + 3a + 3b & = 3x - 2 \end{aligned}$

Equating the coefficients on both sides, we have: $a = -3$ and $3a+3b = -2$ $\implies b = \dfrac 73$ . Therefore $f(x) = \boxed{\frac 73 - 3x}$ .

Let $f(x)=mx+b$ . The equation can be written as the identity equation $mx+m+b+2m-2mx+2b=3x-2$ , where $m$ is the slope and $b$ is the $y$ -intercept. $m=-3$ and we can plug this back in to solve for $b$ , where $3b-9=-2$ so $b$ is $\frac{7}{3}$ .