Calculate f(3), given that f(x) = x3 + f0(-1)x2 + f00(1)x + f0(-1)f(-1).

Differentiating, we compute
f0(x) = 3x2 + 2f0(-1)x + f00(1)
f00(x) = 6x + 2f0(-1):
Plugging in x = -1 into the first equation and x = 1 into the second, we obtain a system of
equations in f0(-1) and f00(1):
f0(-1) = 3 - 2f0(-1) + f00(1);
f00(1) = 6 + 2f0(-1):
Denoting f0(-1) and f00(1) by a and b respectively, we have
a = 3 - 2a + b
b = 6 + 2a:
Plugging the second equation into the first, we get
a = 3 - 2a + (6 + 2a) = 9:
Thus, b = 6 + 2a = 24, so
f(x) = x3 + ax2 + bx + af(-1) = x3 + 9x2 + 24x + 9f(-1):
Finally, plugging x = -􀀀1 into this equation, we get
f(-1) =-1 + 9-24 + 9f(-1)
-8f(-1) = -16
f(-1) = -2
We conclude that
f(x) = x3 + 9x2 + 24x + 18
f(3) = 33 + 9* 32 + 24 *3 + 18
= 27 + 81 + 72 + 18 = 198
correct answer A