Answer is: option4

Solution:

Find the critical points by computing the first derivative and setting it to zero:

f'(x) = d/dx (2x³ + 3x² − 12x + 4) = 6x² + 6x − 12

Set f'(x) = 0:

6x² + 6x − 12 = 0    ⟹    x² + x − 2 = 0

The solutions are x = 1 and x = −2. Since x = −2 is outside the interval [0, 2], the only critical point in the interval is x = 1.

Evaluate the function at the critical point and the endpoints of the interval:

• At x = 0:
  f(0) = 2(0)³ + 3(0)² − 12(0) + 4 = 4


• At x = 1:
  f(1) = 2(1)³ + 3(1)² − 12(1) + 4 = 2 + 3 − 12 + 4 = −3


• At x = 2:
  f(2) = 2(2)³ + 3(2)² − 12(2) + 4 = 16 + 12 − 24 + 4 = 8

The values at the critical point and endpoints are 4, −3, and 8. The maximum value among these is 8.

Answer: D