Answer
$f(x) = \frac{1}{2}x^2-x^3+4x^4+2x+1$
Work Step by Step
$f''(x) = 1-6x+48x^2$
$f'(x) = \int ( 1-6x+48x^2)~dx = x-3x^2+16x^3+C_1$
Note that $f'(0) = 2$
We can find the value of $C_1$:
$f'(0) = (0)-3(0)^2+16(0)^3+C_1 = 2$
$C_1 = 2$
Then:
$f'(x) = x-3x^2+16x^3+2$
$f(x) = \int (x-3x^2+16x^3+2)~dx = \frac{1}{2}x^2-x^3+4x^4+2x+C_2$
Note that $f(0) = 1$
We can find the value of $C_2$:
$f(0) = \frac{1}{2}(0)^2-(0)^3+4(0)^4+2(0)+C_2 = 1$
$C_2 = 1$
Therefore:
$f(x) = \frac{1}{2}x^2-x^3+4x^4+2x+1$
