Question

Given f^ prime prime (x)=x+2 and f^ prime (0)=3 and f(0) = - 1. Find f(x).

Answer 1

Given

The derivative function is given as

$f^{\prime}^{\prime}(x)=x+2$

and f(0) = - 1 and f'(0) = 3

Explanation

To determine the function,

$\begin{gathered} \int d^2y=\int(x+2)dx^2 \\ \frac{dy}{dx}=\frac{x}{2}^2+2x+C \\ f^{\prime}(0)=\frac{0}{2}^2+2(0)+C \\ 3=C \end{gathered}$

It is also given that f(0) = - 1.

$f^{\prime}(x)=\frac{x^2}{2}+2x+3$

Take the integral and find the function

$\begin{gathered} \int dy=\int\frac{x^2}{2}+2x+3dx \\ y=\frac{x^3}{6}+\frac{2x^2}{2}+3x+C \\ y=\frac{x^3}{6}+x^2+3x+C \\ f(x)=\frac{x^3}{6}+x^2+3x+C \\ -1=0+C \\ C=-1 \end{gathered}$

Then the function is determined as

$y=\frac{x^3}{6}+x^2+3x-1$

Answer

Hence the function is determined as

$f(x)=\frac{x^3}{6}+x^2+3x-1$