Answer
$f(x) = \frac{2}{9}x^3+\frac{1}{3}x^2-\frac{4}{3}x+\frac{7}{9}$
Work Step by Step
$f(x) = ax^3+bx^2+cx+d$
$f'(x) = 3ax^2+2bx+c$
The following is given in the question:
$f'(-2) = f'(1) = 0$
$f(-2) = 3$
$f(1) = 0$
Then:
1. $f'(-2) = 3a(-2)^2+2b(-2)+c = 12a-4b+c = 0$
2. $f'(1) = 3a(1)^2+2b(1)+c = 3a+2b+c = 0$
3. $f(-2) = a(-2)^3+b(-2)^2+c(-2)+d = -8a+4b-2c+d = 3$
4. $f(1) = a(1)^3+b(1)^2+c(1)+d = a+b+c+d = 0$
We can add Equation (1) and $2\times$ Equation (2):
$18a+3c = 0$
$c = -6a$
We can subtract Equation (4) from Equation (3):
$-9a+3b-3c = 3$
$-9a+3b-3(-6a) = 3$
$9a+3b = 3$
$3a+b=1$
$b = 1-3a$
We can add Equation (1) and Equation (3):
$4a-c+d = 3$
$4a-(-6a)+d = 3$
$10a+d = 3$
$d = 3-10a$
We can use Equation (1) to find the value of $a$:
$12a-4b+c = 0$
$12a-4(1-3a)+(-6a) = 0$
$18a-4 = 0$
$a = \frac{2}{9}$
We can find the value of $b$:
$b = 1-3a$
$b = 1-3(\frac{2}{9})$
$b = \frac{1}{3}$
We can find the value of $c$:
$c = -6a$
$c = -6(\frac{2}{9})$
$c = -\frac{4}{3}$
We can find the value of $d$:
$d = 3-10a$
$d = 3-10(\frac{2}{9})$
$d = \frac{7}{9}$
Therefore:
$f(x) = \frac{2}{9}x^3+\frac{1}{3}x^2-\frac{4}{3}x+\frac{7}{9}$
