Question

college calculus i tried it a couple times but i haven't been able to get it. from my understanding you substitute each letter until you get numerical values for each and plug them into the given equation

Answer 1

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