Question 287534
f''(x) = x^2,   

f'(x) = (1/3)x^3+C..... f'(0)=4 ---> 4=(1/3)(0)^3+C ---> C=4 ----> f'(x)=(1/3)x^3+4



f(x) = (1/12)x^4+4x+D


2 = (1/12)(0)^4+4(0)+D



2 = 0+D


D = 2



Answer: f(x) = (1/12)x^4+4x+2