Answer
$f(x) = \frac{1}{4}x^5+\frac{1}{2}x^4+x^2-\frac{27}{4}x+3$
Work Step by Step
$f''(x) = 5x^3+6x^2+2$
$f'(x) = \int (5x^3+6x^2+2)~dx = \frac{5}{4}x^4+2x^3+2x+C_1$
$f(x) = \int (\frac{5}{4}x^4+2x^3+2x+C_1)~dx = \frac{1}{4}x^5+\frac{1}{2}x^4+x^2+C_1x+C_2$
Note that $f(0) = 3$
We can find the value of $C_2$:
$f(0) = \frac{1}{4}(0)^5+\frac{1}{2}(0)^4+(0)^2+C_1(0)+C_2 = 3$
$C_2 = 3$
Note that $f(1) = -2$
We can find the value of $C_1$:
$f(1) = \frac{1}{4}(1)^5+\frac{1}{2}(1)^4+(1)^2+C_1(1)+3 = -2$
$\frac{1}{4}+\frac{1}{2}+1+C_1+3 = -2$
$\frac{19}{4}+C_1 = -2$
$C_1 = -2-\frac{19}{4}$
$C_1 = -\frac{27}{4}$
Therefore:
$f(x) = \frac{1}{4}x^5+\frac{1}{2}x^4+x^2-\frac{27}{4}x+3$
