Question

Find f such that the given conditions are satisfied. f'(x) = x2 + 3, f(0) = 21 f(x) X3 + 3x + 21 3 f(x) = x3 + 3x2 + 21 f(x) = +3 + 3x f(x) = x3 + 3x + 21

Answer 1

We have to find f(x).

First we have to integrate f'(x) and then satisfy the initial condition f(0).

Integrating f'(x):

$f(x)=\int (x^2+3)dx=\frac{x^3}{3}+3x+C$

We can then replace x with x=0 and f(x) with f(0)=21 to find the value of the constant C:

$\begin{gathered} f(0)=21=\frac{0^3}{3}+3\cdot0+C \\ C=21 \end{gathered}$

Then, the function f(x) is:

$f(x)=\frac{x^3}{3}+3x+21$

Answer: f(x) = x^3/3 + 3x + 21 (Option A)