Answer
$f(x)=4x^3-5x-3$
Work Step by Step
Integrate $f''(x)$:
$\int f''(x) dx=\int (24x)dx=12x^2+C$
$f'(x)=12x^2+C$
Set in $f'(-1)=7$ and solve for $C$:
$7=12(-1)^2+C$
$C=-5$
$f'(x)=12x^2-5$
Integrate $f'(x)$:
$\int f'(x) dx=\int (12x^2-5)dx=4x^3-5x+C$
$f(x)=4x^3-5x+C$
Set in $f(1)=-4$ and solve for $C$:
$-4=4(1)^3-5(1)+C$
$-4=-1+C$
$C=-3$
$f(x)=4x^3-5x-3$
