f(x) = ax³ + bx² + cx + d.
1) Maximum(-2,4)
2) Minimum(1,0)
The critical points, Max or Min when the derivative f ' (x) = 0
f ' (x) = 3ax² + 2bx + c = 0
a) Maximum(-2,4) that means:
f(-2) = 4 and
f ' (-2) = 0
b) Minimum(1,0) that means:
f(1) = 0 and
f ' (1) = 0
Now plug the values in the related equation:
f(-2) = 4 ↔↔ - 8a + 4b - 2c + d = 4
f ' (-2) = 0 ↔↔ +12a - 4b + c = 0
f(1) = 0 ↔↔ a + b + c + d =0
f ' (1) = 0 ↔↔ 3a + 2b + c = 0
Solving this equation gives:
a= 8/27 , b= 4/9, c=-16/9, d=28/27
f(x) = (8/27)x³ + (4/9)x² - (16/9)x + 28/27
