Find a cubic function, in the form f (x) = ax^3+bx^2+cx+d, that has a local maximum value of 4 at - 3 and a local minimum value of 0 at 2. f (x) = ax3 + bx2 + cx + d.

Wea re given with f (x) = ax^3+bx^2+cx+d

taking the first derivative

f′ (x) = 3ax2 + 2bx + c

substituting the conditions to the original equation

f (-3) = 4=> - 27a + 9b-3c+d

f (2) = 0=>8a+4b+2c+d

f′ (-3) = 0=>27a-6b + c

f′ (2) = 0=>12a+4b+c

solving through the 4x4 solver

a = 8/125

b=12/125

c=-1.152

d=1.408

The equation is equal to 8/125 x^3 + 12/125 x^2 - 1.152 x + 1.408.