The function f has a local extreme at point (1,4). If f''(x)=3x^2+2x, then find f(0)?

A local extreme point would mean that f'(1)=0. Since integration is the inverse application of derivation, to obtain f'(x) function we integrate the f''(x) function. Bearing in mind the integration rule of increasing the power "n" to "n+1" and dividing the new number by "n+1", the integration of f''(x) would yield f'(x)=x^3+x^2+C (C is the constant generated by integrating and can be any constant value. (Remember, after derivation all constants are lost so we can not be certain whether such value of C is 0 or any other real number). As we know that f'(1)=0 (see first sentence), replacing x with 1 and solving the equation would lead to C=-2; thus, f'(x)= x^3+x^2-2.As we now have the f'(x) function, we can use similar integration methods to obtain f(x) function. Hence, f(x)=(x^4)/4+(x^3)/3-2x+C. The question itself provided that f(1)=4. Plugging in these values would give f(1)=1/4+1/3-2+C=4. Solving this, we obtain C=65/12, and f(x)=1/4+1/3-2+65/12. Finally, using this equation when x=0 , we get f(0)=65/12.