Answer
$f(x)$ = $\frac{1}{9}(2x^{3}+3x^{2}-12x+7)$
Work Step by Step
$f(x)$ = $ax^{3}+bx^{2}+cx+d$
$f'(x)$ = $3ax^{2}+2bx+c$
we are given that
$f(1)$ = $0$ and $f(-2)$ = $3$ so
$f(1)$ = $a+b+c+d$ = $0$
$f(-2)$ = $-8a+4b-2c+d$ = $3$
$f'(1)$ = $3a+2b+c$ = $0$
$f'(-2)$ = $12a-4b+c$ = $0$
Solve 4 equations we get
$a$ = $\frac{2}{9}$, $b$ = $\frac{1}{3}$, $c$ = $-\frac{4}{3}$, $d$ = $\frac{7}{9}$
so function is
$f(x)$ = $\frac{1}{9}(2x^{3}+3x^{2}-12x+7)$
