2y3+3y2-2y+1 найдите f(x), f(0), f(-1), f(2x-1)

Если  f(y)=2y³+3y²-2y+1, то
f(x)=2x³+3x²-2x+1
f(0)=2·0³+3·0²-2·0+1=1
f(-1)=2·(-1)³+3·(-1)²-2·(-1)+1=-2+3+2+1=4
 f(2x-1)= 2·(2x-1)³+3·(2x-1)²-2·(2x-1)+1=2·(8х³-12х²+6х-1)+3·(4х²-4х+1)-4х+2+1=
=16х³-24х²+12х-2+12х²-12х+3-4х+3=16х³-12х²-4х+4