Answer
$g(x)=\frac{4}{3}x^3+\frac{13}{3}$
Work Step by Step
First, we integrate.
$g'(x)=4x^2\hspace{10mm}g(-1)=3$
Check: $\int g'(x)dx=\int4x^2dx=\frac{4}{3}x^3+C=g(x)$
Now, we find $C$ using the given point:
$g(-1)=\frac{4}{3}(-1)^3+C=-\frac{4}{3}+C=3$
$C=3+\frac{4}{3}=\frac{13}{3}$
$g(x)=\frac{4}{3}x^3+\frac{13}{3}$
