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· (8 х³-12 х²+6 х-1) + 3· (4 х²-4 х+1) - 4 х+2+1=

=16 х³-24 х²+12 х-2+12 х²-12 х+3-4 х+3=16 х³-12 х²-4 х+4