Answer
$$f\left( x \right) = {x^2} - 3x + 4$$
Work Step by Step
$$\eqalign{
& f'\left( x \right) = 2x - 3 \cr
& f\left( x \right) = \int {f'\left( x \right)} dx \cr
& then \cr
& f\left( x \right) = \int {\left( {2x - 3} \right)} dx \cr
& find{\text{ the general solution}} \cr
& f\left( x \right) = {x^2} - 3x + C \cr
& {\text{using the initial condition }}F\left( 0 \right) = 4 \cr
& 4 = {\left( 0 \right)^2} - 3\left( 0 \right) + C \cr
& C = 4 \cr
& {\text{the solution to the initial value problem is}} \cr
& f\left( x \right) = {x^2} - 3x + 4 \cr} $$
