If $X$~$Uniform(0,1)$ then $-2X+1$~$Uniform(-1,1)$
This statement is true. But I'm not sure the way I approach this problem is necessarily correct.
I would start with solving $E[-2X+1]$, which boils down to $-2E[X]+1 = -.5$ EDIT: Error here, $-2E[X]+1 = 0$
Then I would call $Y$~$Uniform(-1,1)$, and find the $E[Y]$, which is $0$.
At least, that's how I would approach the problem. Is this enough work?
You can calculate the cdf. Let $Z=-2X+1$. Then
$P(Z<z)=P(-2X+1<z)=P(1-z<2X)=P\left(\frac{1-z}{2}<X \right)$
The cdf of X is
$$F_X(x)=\begin{cases}0, \text{if} \ x<0 \\ x, \text{if} \ 0 \leq x<1 \\ 1, \text{if} \ x\geq 1 \end{cases}$$
Then use $P\left(\frac{1-z}{2}<X \right)=1-P\left(X<\frac{1-z}{2} \right)$. And you have to adjust the limits: $x=0 \Rightarrow z=1, x=1\Rightarrow z=-1$
Therefore the cdf of $Z$ is
$$F_Z(z)=\begin{cases}0, \text{if} \ z<-1 \\ 1-\frac{1-z}{2}, \text{if} \ -1 \leq z<1 \\ 1, \text{if} \ z\geq 1 \end{cases}$$
Is Z uniformly distributed on $(-1,1)$ ?