Given
The derivative function is given as
[tex]f^{\prime}^{\prime}(x)=x+2[/tex]
and f(0) = - 1 and f'(0) = 3
Explanation
To determine the function,
[tex]\begin{gathered} \int d^2y=\int(x+2)dx^2 \\ \frac{dy}{dx}=\frac{x}{2}^2+2x+C \\ f^{\prime}(0)=\frac{0}{2}^2+2(0)+C \\ 3=C \end{gathered}[/tex]
It is also given that f(0) = - 1.
[tex]f^{\prime}(x)=\frac{x^2}{2}+2x+3[/tex]
Take the integral and find the function
[tex]\begin{gathered} \int dy=\int\frac{x^2}{2}+2x+3dx \\ y=\frac{x^3}{6}+\frac{2x^2}{2}+3x+C \\ y=\frac{x^3}{6}+x^2+3x+C \\ f(x)=\frac{x^3}{6}+x^2+3x+C \\ -1=0+C \\ C=-1 \end{gathered}[/tex]
Then the function is determined as
[tex]y=\frac{x^3}{6}+x^2+3x-1[/tex]
Answer
Hence the function is determined as
[tex]f(x)=\frac{x^3}{6}+x^2+3x-1[/tex]
