find two possible functions f given the second- or third-order derivatives.

Question

f " (x)=1+x

please try to do it step by step

Answer 1

y'' = 1 + x

integrates:

y' = x + (1/2)x^2 + c

integrates again:

y = (1/2)x^2 + (1/6)x^3 + cx + k

c and k are fixed # constants

check by twice differentiating:

D^2 (1/2)x^2 = D [D (1/2)x^2 ] = D [ x] = 1

D^2 (1/6)x^3 = D [ D (1/6)x^3] = D (1/2 x^2) = x

cx+k vanishes, leaving only 1+x

y = f(x) = (1/6)x^3 + (1/2)x^2 + cx + k