## Calculus: Early Transcendentals 8th Edition

$f(x)=-x^{2}+2x^{3}-x^{4}+12x+4$
$f''(t)=-2+12x-12x^{2}$ 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 $D=4$ $f(x)=-x^{2}+2x^{3}-x^{4}+12x+4$