Найдите f '(1) + f(1), если f(х) =
[tex]x ^{2} (3x^{2} - 2)[/tex]

f'(x)= 2x(3x²-2) · 6x³

f'(1)= 12

f(1)= 1

f'(1)+f(1)= 12+1= 13

f'(x) = (x²(3x²-2))' = (3x⁴ - 2x²)' = 12x³ - 4x

f '(1) + f(1) = 12*1³ - 4*1 + 1²(3*1²-2) = 12 - 4 + 3 - 2 = 9