Answer
$f(x)=-x^{2}+2x^{3}-x^{4}+12x+4$
Work Step by Step
$f''(t)=-2+12x-12x^{2}$
Using the antiderivatives table,
$f'(x)=-2x+12\displaystyle \cdot\frac{x^{2}}{2}-12\cdot\frac{x^{3}}{3}+C$
$f'(x)=-2x+6x^{2}-4x^{3}+C$
Given that $f'(0)=12$,
it follows that $C=12,$
$f'(x)=-2x+6x^{2}-4x^{3}+12$
Using the antiderivatives table,
$f(x)=-2\displaystyle \cdot\frac{x^{2}}{2}+6\cdot\frac{x^{3}}{3}-4\cdot\frac{x^{4}}{4}+12x+D$
$f(x)=-x^{2}+2x^{3}-x^{4}+12x+D$
Given that $f(0)=4$, it follows that $D=4$
$f(x)=-x^{2}+2x^{3}-x^{4}+12x+4$
